On the Question of Approximate Methods for Solving Equations of the Third Degree

A. S. Begekhtin

Introduction

In the investigation of certain questions of a theoretical and engineering nature, the need often arises to solve equations of the third degree.

An exact and comparatively rapid solution of a cubic equation is possible only when, by means of suitable transformations, it can be represented either in the form \((x \pm a)^3 = 0\), or in the form \((x \pm a)^3 = m\), or, finally, in the form \((x - a_1)(x - b_1)(x - c_1) = 0\). In the remaining cases one must be satisfied with approximate solutions with one or another degree of accuracy.

Before turning to the question under consideration, it is useful, if only briefly, to dwell on the generally known methods for determining, with a prescribed degree of accuracy, roots of the second and third degrees, as well as roots of algebraic equations, since these methods, to a certain extent—chiefly as control methods—may also be used in the method we propose for solving equations of the third degree.

The method of extracting square and cubic roots, discovered by Indian mathematicians, was almost unknown in Europe until the end of the sixteenth century¹. Only at the end of the sixteenth century, through the Arabs, probably from Spain, did it penetrate into Europe. In 1600 this method was developed by the mathematician Vieta, who adapted it to the computation of roots of algebraic equations. But in the form in which it was proposed by Vieta, this method proved in practice to be so laborious that one seventeenth-century mathematician called the computation of roots of algebraic equations by means of it “work unworthy of a Christian.” Nevertheless, it persisted for about 80 years, approximately until 1680.

In 1674 Gregory and Darri almost simultaneously and independently of one another developed a method for determining square roots, called the “principle of iteration.” This principle was subsequently developed by Newton for determining the roots of algebraic equations.

As applied to the extraction of square roots, the principle of iteration reduces to the following:

“Let there be a number \(N\), and suppose it is required to find its square root. Take some number \(x_0\), and with its help form the number \(x_1\), satisfying the equation:
\[ x_1 = \frac{1}{2}\left(x_0 + \frac{N}{x_0}\right). \]
From \(x_1\) we likewise form the number \(x_2\),

satisfying the equation \(x_2=\dfrac{1}{2}\left(x_1+\dfrac{N}{x_1}\right)\). From \(x_2\) we form the equation \(x_3=\dfrac{1}{2}\left(x_2+\dfrac{N}{x_2}\right)\), and so on. Then the sequence of numbers \(x_0,\ x_1,\ x_2,\ x_3,\ldots\) tends to the limit \(\sqrt{N}\)*).

Example. Find \(\sqrt{10}\).

Solution

Putting \(N=10\) and \(x_0=1\), we have

\[ x_1=\frac{1}{2}\left(1+\frac{10}{1}\right)=5.5, \]

\[ x_2=\frac{1}{2}\left(5.5+\frac{10}{5.5}\right)=\frac{1}{2}(5.5+1.8)=3.7, \]

\[ x_3=\frac{1}{2}\left(3.7+\frac{10}{3.7}\right)=\frac{1}{2}(3.7+2.7)=3.2, \]

\[ x_4=\frac{1}{2}\left(3.2+\frac{10}{3.2}\right)=\frac{1}{2}(3.2+3.125)=3.163, \]

\[ x_5=\frac{1}{2}\left(3.163+\frac{10}{3.163}\right)=\frac{1}{2}(3.163+3.161555)=3.1622775. \]

The last number is \(\sqrt{10}\) with accuracy to the 6th decimal place. It is not hard to see that for another value of \(x_0\) an analogous result will also be obtained. Thus, for \(x_0=2\), \(x_1=3.5\), \(x_2=3.18\), \(x_3=3.16\), and \(x_4=3.16228\).

The Newton–Raphson method, proposed by Newton (1685) and Raphson (1690) for computing roots of algebraic equations, amounts to the following.

If \(x_0, x_1, x_2, x_3,\ldots\) are approximate values successively tending to the true value of the root of \(f(x)=0\), then between \(x_0, x_1, x_2\), etc., there must exist the following relation:

\[ x_1=x_0-\frac{f(x_0)}{f'(x_0)};\qquad x_2=x_1-\frac{f(x_1)}{f'(x_1)} \]

and, in general,

\[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. \]

Example. Find the positive root of the equation \(x^3-2x-5=0\).

Solution. The first approximate value of the root of this equation will obviously be \(x_0=2\). It is less than the true one, since \(2^3<2\cdot2+5\).

\[ x_1=2-\frac{f(2)}{f'(2)}=2+\frac{1}{10}=2.1, \]

\[ x_2=2.1-\frac{f(2.1)}{f'(2.1)} =2.1-\frac{0.061}{11.23} =2.1-0.0054=2.0946 \]

with accuracy to the third decimal place.

) Borrowed from the book by E. Whittaker and R. Robinson, Mathematical Processing of Observational Results*, ONTI, 1935, pp. 78–79, where a proof of this proposition is given.

The accuracy with which the value of \(x\) has been found can be determined by means of a method called the “rule of false position.” Its essence reduces to the following.

If \(a, a_1, a_2\) and \(b, b_1, b_2\) are approximate values of the root of the equation \(f(x)=0\), with the values \(a, a_1, a_2\) less than, and the values \(b, b_1, b_2\) greater than, the true values of \(x\), and if the subsequent values \(a, a_1, a_2\) and \(b, b_1, b_2\) are expressed in terms of the preceding ones as follows:

\[ a_1=a-\frac{(b-a)f(a)}{f(b)-f(a)}, \qquad a_2=a_1-\frac{(b_1-a_1)f(a_1)}{f(b_1)-f(a_1)}, \]

\[ b_1=b-\frac{f(b)}{f'(b)}, \qquad b_2=b_1-\frac{f(b_1)}{f'(b_1)}, \]

then the degree of accuracy in computing \(x\) will be determined by the number of identical decimal digits in \(a\) and \(b\), \(a_1\) and \(b_1\), \(a_2\) and \(b_2\).

For example, the first approximate (underestimated) value of the root of the equation \(x^3-23.4x-8=0\) will be
\(a=\sqrt{23.4}=4.793\), and the second (overestimated)

\[ b=\sqrt{23.4+\frac{8}{4.793}}=5.006,\quad b-a=5.006-4.793=0.213, \]

\[ f(a)=-9.05,\quad f(b)=+0.02,\quad a_1=4.793-\frac{0.213(-9.05)}{9.07}=5.0055,\quad f'(b)= \]

\[ =44.58,\quad b_1=5.006-\frac{0.02}{44.58}=5.0056. \]

The number of identical decimal digits in \(a_1\) and \(b_1\) is three; that is, for the given stage of calculation the root of the equation has been found with accuracy to thousandths. The true value is \(x=5\), the error \(0.1\%\).

Turning to the problem we have set, it should be noted that, besides Cardano’s classical method, which does not always give correct solutions,\(^{1}\) there exist quite a number of methods for the approximate solution of equations of the third degree: the methods of Newton, Graeffe, Encke, the “method of false position,” and so on. There is even a work,\(^{2}\) using which one can, from the coefficients with the unknown (assuming that the constant term is the coefficient of \(x^0\)), find all three roots of the equation.

Without entering into a comparative evaluation of these methods, we shall merely indicate that determining by means of them the value of the unknown with a sufficient degree of accuracy is a rather cumbersome and laborious process. As for the above-mentioned work by Engineer V. S. Osipov, it can be used to obtain values of \(x\) lying within definite, comparatively narrow limits.

The methods proposed below for the approximate solution of equations of the third degree, being simple and uncomplicated, make it possible comparatively easily and quickly to determine the value of the unknown with the desired degree of accuracy. They may be applied to the solution of any type of equation of the third degree, provided only that this equation has at least one real root—whether positive or negative.

\(^{1}\) Military engineer of the 3rd rank V. S. Osipov. Tables for solving cubic equations. Military Engineering Academy of the Red Army named after V. V. Kuibyshev, Moscow, 1937, p. 3.

\(^{2}\) Ibid.

§ 1. Approximate solution of an incomplete equation of the third degree of the form

\[ x^3-bx-c=0 \]

Incomplete equations of the third degree may be divided into 8 groups, differing from one another either by the signs of \(x\) and \(x^2\) when the sign of the constant term is the same, or by the signs of the constant term when the signs of \(x\) and \(x^2\) are the same.

\[ \begin{aligned} 1.\;& x^3-bx-c=0 &\qquad 5.\;& x^3-ax^2-c=0\\ 2.\;& x^3-bx+c=0 &\qquad 6.\;& x^3-ax^2+c=0\\ 3.\;& x^3+bx-c=0 &\qquad 7.\;& x^3+ax^2-c=0\\ 4.\;& x^3+bx+c=0 &\qquad 8.\;& x^3+ax^2+c=0 \end{aligned} \]

There is no need to investigate all 8 groups: an equation provisionally assigned by us to the fourth group, by a simple replacement of \(c\) by \(-c\), is transformed into an equation of the third group, while the equations of the last four groups may be reduced to equations of one of the first three groups by replacing \(x\) by \(y \pm \dfrac{a}{3}\).

We proceed to solve the equation provisionally assigned to the first group, i.e., the equation \(x^3-bx-c=0\).

Let there be the equation

\[ x^3-b_1x-c=0, \tag{1} \]

for which \(x=2z\), where \(z=\sqrt[3]{c}\). If this condition is satisfied, then equation (1) may be represented in the form: \((2z)^3-b_1 2z-z^3=0\), or

\[ 2zb_1=7z^3, \tag{2} \]

whence

\[ b_1=\frac{7z^3}{2z}=3.5z^2 \tag{3} \]

Thus, the condition \(x=2\sqrt[3]{c}\) can be satisfied in the case when the coefficient of the unknown in the first degree is \(b_1=3.5\sqrt[3]{c^2}\). Obviously, in practice such relations among \(x\), \(b_1\), and \(c\) can occur only in an exceptional case.

In order to use these simple relations for solving the equation

\[ x^3-bx-c=0 \tag{1a} \]

one may proceed in two ways.

Method I. Leaving unchanged the value of \(c\) in equation (1a), find what the coefficient \(b\) in this equation must be in order that one may take

\[ x=x_1=2z=2\sqrt[3]{c}. \]

As has already been found above, in this case the value of \(b\) must be \(3.5z^2\), or \(3.5\sqrt[3]{c^2}\). Usually, as a rule, \(b \ne 3.5z^2\). If in equation (1a) \(3.5z^2>b\), then this will mean that the value \(x=2z\) will be greater than the true one, since, in order to compensate the quantity \(x^3-c\), it is necessary to subtract from it not \(bx\), as required by equation (1a), but \(3.5z^2x\), a quantity larger than \(bx\). If, however, \(3.5z^2<b\), the value \(x=2z\) will obviously be less than the true one, i.e., it will be found with a deficiency.

If \(d\)—the difference between the true value \(x\) and \(x_1\)—is equal to \(2z\), then, substituting in equation (1a) \(x_1 \pm d\) instead of \(x\) and solving it with respect to \(d\), we obtain the value of the difference between \(x_1\) and \(x\). In doing so, the quantity \(d^3\), as being too small in comparison with the other quantities of the expanded equation \((x_1+d)^3-b(x_1+d)-c=0\), may be neglected, so that the determination of \(d\) reduces to the simple solution of a quadratic equation. It is still better to solve simultaneously the equations \(x^3-bx-c=0\) and \(x_1^3-b_1x_1-c=0\) (where \(b_1=\)

\[ =3.5\sqrt[3]{c^2}), \]

since in this case \(x_1^3\) and \(c\) disappear.

If \(x_1>x\), then the joint solution of both equations gives:

\[ \begin{aligned} &x_1^3-3x_1^2d+3x_1d^2-d^3-bx_1+bd-c=0\\ -&\frac{x_1^3\qquad\qquad\qquad\quad -bx_1\qquad -c=0}{3x_1d^2-(3x_1^2-b)d+(b_1-b)x_1=0} \end{aligned} \]

or

\[ d^2-\left(x_1-\frac{b}{3x_1}\right)d+\frac{b_1-b}{3}=0, \tag{4} \]

whence \(d\) is easily determined, since the quantities \(x_1\), \(b\), and \(b_1\) may be regarded as known.

If \(x_1<x\), then equation (4) takes the form

\[ d^2+\left(x_1-\frac{b}{3x_1}\right)d+\frac{b_1-b}{3}=0, \tag{5} \]

i.e., it will differ from the preceding one only in the sign before \(d\).

Method II. Leaving \(b\) unchanged, from the expression \(b=3.5\sqrt[3]{c_1^2}\) we determine \(c_1\), i.e., the value of the constant term for which \(x_2=2z_2=2\sqrt[3]{c_1}\). If \(c_1<c\), then \(x_2=2z_2\) will be less than the true value, since, in order to satisfy the condition \(x^3-bx-c=0\), one must subtract from \(x^3-bx\) a quantity \(c_1\) smaller than that required by equation (1a). In the opposite case, i.e. when \(c_1>c\), \(x_2>x\).

Subsequently, the determination of the difference between \(x\) and \(x_2\) is carried out in the same way as in the preceding case.

Example. Find the real root of the equation \(x^3-23.4x-8=0\).

Solution. In the present case \(b=23.4\) and \(c=8\).

We find:

\[ b_1=3.5z_1^2=3.5\sqrt[3]{8^2}=14;\quad z_1=\sqrt[3]{8}=2;\quad x_1=2z_1=4. \]

Since \(b_1<b\), we have \(x_1<x\).

We determine \(d\) from expression (5):

\[ d^2+\left(4-\frac{23.4}{3\cdot4}\right)d+\frac{14-23.4}{3}=0, \]

or

\[ d^2+2.05d-3.133=0, \]

whence

\[ d=-1.025+\sqrt{1.025^2+3.133}=-1.025+2.045=1.02. \]

Thus, \(x-x_1=1.02\), whence \(x=4.00+1.02=5.02\).

To refine the value of \(x\) found, one may apply the Newton–Raphson rule. According to this rule,

\[ x_y=x-\frac{f(x)}{f'(x)}. \]

In the example under consideration \(x=5.02\); \(f(x)\), as the computations show, is \(1.038\); \(f'(x)=52.2\), and the refined value \(x=x_y\) will be:

\[ x_y=5.02-\frac{1.038}{52.2}=5.02-0.0199=5.0001. \]

Let us solve the given equation \(x^3-23.4x-8=0\), leaving \(b\) unchanged and determining \(c_1\). For the example under consideration,

\[ c_1=\sqrt{\left(\frac{23.4}{3.5}\right)^3}=17.28 \]

and

\[ z_2=\sqrt[3]{17.28}=2.585, \]

whence

\[ x_2=2z_2=2\cdot 2.585=5.17. \]

Since \(c_1>c\), it follows that \(x_2>x\).

Determine \(d_1\) from the equation \((x_2-d_1)^3-23.4(x_2-d_1)-8=0\), which, after the corresponding transformations, takes the form:

\[ d_1^2-\left(x_2-\frac{b}{3x_2}\right)d_1+\frac{c_1-c}{3x_2}=0 \]

or

\[ d_1^2-3.66d_1+0.597=0, \]

whence

\[ d_1=1.83-\sqrt{1.83^2-0.597}=1.83-1.66=0.17. \]

Thus,

\[ x=x_2-d_1=5.17-0.17=5.00. \]

Obviously, the value \(x=5\) is closer to the true value than \(x=5.02\). Indeed, in the first case, i.e., for \(x=5.02\), the difference between the roughly approximated value \(x=x_1=4\) and its true value \((5)\) amounts to \(20\%\) of the true value of \(x\). Therefore neglecting the quantity \(d^3\) gave an error of \(0.02\), or \(0.4\%\). In the second case, i.e., for \(x=x_2=5.17\), \(x_2-x=0.17\), or \(3.4\%\) of the value of \(x\), and neglecting the quantity \(0.17^3\) did not affect the final result. Thus, \(x=5.00\).

The other two roots of the given equation are determined from the condition that their sum, taken with the opposite sign, is equal to \(x\), and their product to \(\dfrac{c}{x}\).

In short, they are determined from the quadratic equation \(x^2+5x+1.6=0\). Its solution gives:

\[ x'=-2.5+\sqrt{6.25-1.6}=-2.5+\sqrt{4.65}=-0.34, \]

\[ x''=-2.5-\sqrt{4.65}=-2.5-2.16=-4.66. \]

§ 2. Approximate solution of an equation of the form

\[ x^3-bx+c=0. \]

Let us find the values of the coefficient of \(x\), or of the constant term, for which the condition

\[ x_1=2z_1=2\sqrt[3]{c} \]

or

\[ x_2=2z_2=\sqrt[3]{c_2} \]

will be satisfied.

Taking in the general case

\[ x=2z=2\sqrt[3]{c}, \]

we shall have:

\[ 8z^3-b\cdot 2z+z^3=0. \]

or

\[ b_1=b=4{,}5\,z^2=4{,}5\sqrt[3]{c^2}. \tag{6} \]

From the last equation we obtain, replacing \(c\) by the required \(c_2\):

\[ b=4{,}5\sqrt[3]{c_2^2}, \]

whence

\[ c_2=\sqrt{\left(\frac{b}{4{,}5}\right)^3}. \tag{7} \]

In all other respects the method of solving the equation under consideration remains the same as in the case of the equation \(x^3-bx-c=0\).

Example. Find the roots of the equation \(x^3-42x+36=0\).

Solution (first variant). We find the value of the coefficient \(b_1\) for which

\[ x_1=2\sqrt[3]{c} \]

\[ b_1=4{,}5\sqrt[3]{(36)^2}=49{,}06. \]

Since \(b_1>b\), \(x_1=2\sqrt[3]{c}\) will be greater than the true value. We find \(z_1=\sqrt[3]{36}=3{,}302\) and \(x_1=2z_1=6{,}604\).

Determine the difference between \(x_1\) and \(x\) from expression (4):

\[ d_1^2-\left(6{,}604-\frac{42}{3\cdot 6{,}604}\right)d_1+\frac{49{,}06-42}{3}=0 \]

or

\[ d_1^2-4{,}484d_1+2{,}353=0, \]

whence

\[ d_1=2{,}242-\sqrt{5{,}024-2{,}353}=2{,}242-1{,}634=0{,}608. \]

Thus, \(x_1=6{,}604-0{,}608=5{,}996\).

Second variant. We find the value \(c_2\) for which \(x_2=2\sqrt[3]{c_2}\):

\[ c_2=\sqrt{\left(\frac{42}{4{,}5}\right)^3}=28{,}52 \]

and

\[ x_2=2\sqrt[3]{28{,}52}=2\cdot 3{,}055=6{,}11. \]

Since, in order for the condition \(x^3-42x+36=0\) to be satisfied at \(x_2=6{,}11=x\), one must add to \(x^3-42x\) not 36, but only 28.52, a quantity smaller than \(c\), it is evident that \(x_2\) is also greater than \(x\).

We find the difference between \(x_2\) and the presumed true value \(x\), for which we use the equation

\[ d_2^2-\left(x_2-\frac{b}{3x_2}\right)d_2+\frac{c-c_2}{3x_2}=0, \tag{8} \]

obtained by jointly solving the equations

\[ (x_2-d_2)^3-b(x_2-d_2)+c=0 \quad\text{and}\quad x_2^3-bx_2+c_2=0. \]

In the case under consideration \(x_2=6{,}11\), \(b=42\), \(c=36\), and \(c_2=28{,}52\), and equation (8) takes the form:

\[ d_2^2-\left(6{,}11-\frac{42}{3\cdot 6{,}11}\right)d_2+\frac{36-28{,}52}{3\cdot 6{,}11}=0, \]

or

\[ d_2^2-3.819d_2+0.4081=0, \]
whence

\[ d_2=1.9095-\sqrt{3.6475-0.4081}=1.9095-1.7999=0.1096. \]

Hence \(x=6.11-0.1096=6.0004\).

Thus, \(x\) lies within the limits \(5.996 \div 6.0004\).

Since in the first variant the relative error, equal to \(0.608\), is almost 6 times greater than in the second, the value closest to the true one will be \(6.0004\). The actual value of \(x\) will be \(6.0\), as is not difficult to verify by substituting this value into the equation \(x^3-42x+36=0\). But even the value of \(x\) obtained in the first variant differs from the true one by only \(0.004\), or by \(0.066\%\). This is an accuracy quite sufficient for technical calculations.

It would have been possible, of course, not to resort to the second variant, but to refine the value of \(x\) obtained in the first variant by applying the Newton–Raphson method. Then we would obtain: \(f(x)=-0.265\), \(f'(x)=65.855\), and

\[ x_y=5.996+\frac{0.265}{65.855}=5.996+0.00402. \]

Thus, \(x=6.00002\). Obviously, for calculations the value \(x=6.0\) would be adopted.

The other roots of the equation are determined from the equation \(x^2+6x-6=0\), which is obtained if \(x^3-42x+36=0\) is divided by \(x-6\). These roots will be: \(x'=-3+\sqrt{15}\) and \(x''=-3-\sqrt{15}\).

§ 3. Approximate solution of an equation of the form

\[ x^3+bx-c=0. \]

In the present case it would be inexpedient to take \(x_1=2\sqrt[3]{c}\), since \(c\) compensates \(x^3+bx\); therefore the value \(x_1\) would turn out to be greatly overestimated, and the accuracy of the solution would be small, since the difference \(x_1-x\) would be obtained with a considerable error. Under these conditions it is more correct to take the coefficient of \(\sqrt[3]{c}\) not as 2, but as less than 1, for example, \(0.9\), i.e., to take

\[ x_1=0.9\sqrt[3]{c}. \tag{9} \]

Then the equation \(x^3+bx-c=0\) will take the form:

\[ 0.729z^3+0.9bz-z^3=0. \tag{10} \]

where \(z=\sqrt[3]{c}\).

From the preceding equation it follows that, in order to satisfy the condition

\[ x_1=0.9\sqrt[3]{c}, \]

it is necessary that

\[ b=b_1=\frac{z^3-0.729z^3}{0.9z}=0.3z^2. \tag{11} \]

Thus, \(b_1=0.3z^2\), or more precisely \(b_1=0.3011z^2\). If \(b_1>b\), then \(x_1<x\), since in this case a larger quantity must be added to \(x^3-c\) in order to satisfy the condition \(x_1^3+bx_1-c=0\).

If the value of the coefficient \(b\) is left unchanged, then the value of the free term \(c_2\) for which the condition \(x_2=0.9\sqrt[3]{c_2}\) will be satisfied is determined from expression (11) for \(b=b_1\):

\[ b=0.3\sqrt[3]{c_2^{\,2}}, \]

whence

\[ c_2=\sqrt{\left(\frac{b}{0.3}\right)^3} \tag{12} \]

If \(c_2<c\), then \(x_2=0.9\sqrt[3]{c_2}<x\), since in this case a quantity smaller than \(c\) must be subtracted from \(x^3+bx\) in order to satisfy the condition \(x^3+bx-c=0\).

Thereafter the course of the solution remains the same as in the preceding cases: 1) either \(b_1\) is determined for the true value of \(c\), \(x_1=0.9\sqrt[3]{c}\) is computed, and the difference between \(x\) and \(x_1\), or \(x_1\) and \(x\), is computed from the expression

\[ d_1^2 \pm \left(x_1+\frac{b}{3x_1}\right)d_1+\frac{b-b_1}{3}=0, \tag{13} \]

where the plus sign before \(d_1\) is taken in the case \(x_1<x\), and the minus sign when \(x_1>x\); 2) or \(c_2\) and \(x_2=0.9\sqrt[3]{c_2}\) are determined, and the difference \(d_2\) is likewise found from the expression

\[ d_2^2+\left(x_2+\frac{b}{3x_2}\right)d_2-\frac{c-c_2}{3x_2}=0, \tag{14} \]

where the plus sign of the coefficient of \(d_2\) is taken when \(x_2<x\), and the minus sign in the opposite case.

Example. Solve the equation \(x^3+5x-84=0\).

Solution (first variant). Determine \(b_1\) and \(x_1\):

\[ b_1=0.3011z^2,\quad \text{or}\quad b_1=0.3011\sqrt[3]{84^2}=5.7753. \]

\[ z_1=\sqrt[3]{84}=4.38\quad \text{and}\quad x_1=0.9\cdot 4.38=3.942. \]

Since \(b_1>b\), we have \(x_1<x\).

Find \(d_1\) from expression (13), taking into account that \(x_1<x\):

\[ d_1^2+\left(x_1+\frac{b}{3x_1}\right)d_1+\frac{b-b_1}{3}=0, \]

or

\[ d_1^2+\left(3.942+\frac{5}{3\cdot 3.942}\right)d_1+\frac{5-5.7753}{3}=0; \]

whence

\[ d_1^2+4.360d_1-0.2584=0, \]

\[ d_1=-2.180+\sqrt{4.7520+0.2584}=-2.180+2.238=0.058. \]

The value of \(x\) will be:

\[ 3.942+0.058=4.000. \]

Second variant of the solution. Determine \(c_2\), for which

\[ z_2=\sqrt[3]{c_2},\quad \text{and}\quad x_2=0.9z_2: \]

\[ c_2=\sqrt{\left(\frac{5}{0.3011}\right)^3}=67.668 \]

\[ z_2=\sqrt[3]{67.668}=4.075, \]

\[ x_2=0.9\cdot 4.075=3.6675. \]

Since \(c_2<c\), we have \(x_2<x\).

We find the correction \(d_2\) from expression (14), taking into account that \(x_2<x\):

\[ d_2^2+\left(3.6675+\frac{5}{3\cdot 3.6675}\right)d_2-\frac{84-67.668}{3\cdot 3.6675}=0, \]

or

\[ d_2^2+4.1219d_2-1.4844=0, \]

whence

\[ d_2=-2.06095+\sqrt{4.2476+1.4844}=-2.06095+2.3942,\quad \text{or } d_2=0.3325. \]

The approximate value of \(x\) in this case will be:

\[ x=3.6675+0.33325=4.00075. \]

Since in the first variant the correction \((0.058)\) is six times smaller than in the second \((0.33325)\), it is obvious that the value \(x\doteq 4.0000\) is closer to the true one than \(x\doteq 4.00075\). Indeed, the true value is \(x=4\).

The two other roots of the equation are determined from the equation \(x^2+4x+21=0\).

They will be: \(x'=-2+\sqrt{-17}\) and \(x''=-2-\sqrt{-17}\).

§ 4. Approximate solution of an equation of the form

\[ x^3+bx+c=0. \]

Equations of this form do not have positive roots, since the coefficients of the unknown and the constant term enter the equation with the same sign—with a plus.

To solve such an equation, one usually changes the sign to the opposite either at the odd powers of \(x\), or at the constant term, and the resulting equation is solved in the usual way, taking into account that the result obtained in solving it must be taken with the minus sign.

When using the method we propose, the solution of an equation of the given form is reduced to the solution of the equation of the form already considered earlier,

\[ x^3+bx-c=0. \]

Example. Solve the equation \(x^3+20x+1200=0\).

Solution. Representing the equation in the form \(x^3+20x-1200=0\), we determine the coefficient \(b_1\), for which

\[ x_1=0.9\sqrt[3]{1200}. \]

According to (11),

\[ b_1=0.3011\sqrt[3]{1200^2},\quad \text{or } \quad b_1=0.3011\cdot 112.92=34.0; \]

hence

\[ x_1=0.9\sqrt[3]{1200}=10.627\cdot 0.9=9.5643. \]

Since \(b_1>b\), we have \(x_1<x\).

The difference \(d_1\) between \(x\) and \(x_1\) is determined from expression (13):

\[ d_1^2+\left(x_1+\frac{b}{3x_1}\right)d_1+\frac{b-b_1}{3}=0, \]

or

\[ d_1^2+\left(9.5643+\frac{20}{28.6929}\right)d_1+\left(\frac{20-34}{3}\right)=0; \]

\[ d_1^2+10.2612d_1-4.6666=0, \]

whence

\[ d_1=-5.1306+\sqrt{26.3177+4.6666} \]

or

\[ d_1=-5.1306+5.5662=0.4356; \]

hence

\[ x=9.5643+0.4356=9.9999. \]

The true value is \(x=10\). The error is \(0.0001\), or \(0.001\%\).

Thus one of the roots of the equation will be \(10\).

Without giving the detailed computations of \(c_2, x_2\), and \(d_2\), we shall only indicate that for the present example \(c_2\) will be \(541.24\), \(x_2=7.3345\), \(d_2=+2.7286\), and \(x=7.3345+2.7286=10.0631\). The error is \(0.0631\), or \(0.63\%\), i.e. 630 times greater than in the first case, although \(d_2\) is only 6 times greater than \(d_1\). Hence one may draw the practically important conclusion that, for the same (or approximately the same) difference between \(b_1, b\) and \(c_2, c\) (in percentage terms), determining \(x\) through \(b_1\) gives considerably more accurate results than through \(c_2\).

§ 5. Approximate solution of an equation of the form

\[ x^3-a x^2-c=0. \]

As was already indicated above, equations of this type, as well as analogous equations containing \(x\) in the second degree, can be transformed into equations of the first three groups by replacing \(x\) by \(y \pm \dfrac{a}{3}\). But they can also be solved by the methods considered above, without resorting to a substitution.

If we assume that \(x=2z=2\sqrt[3]{c}\), then the coefficient of \(x^2\) must have a quite definite value, determined from the equation \((2z)^3-a(2z)^2-z^3=0\), whence it follows that

\[ a=\frac{7}{4}z. \tag{15} \]

In what follows, the method of solution is analogous to that considered earlier.

In particular, one determines \(a_1=\dfrac{7}{4}\sqrt[3]{c}\), establishes whether

\[ x_1=2\sqrt[3]{c} \]

has been obtained with an excess or a deficiency (for \(a_1>a\), \(x_1>x\); for \(a_1<a\), the reverse), and finds the difference between \(x_1\) and \(x\) from the equation

\[ d_1^2 \pm x_1 \frac{(3x_1-2a)}{(3x_1-a)}\, d_1 +\frac{x_1^2(a_1-a)}{3x_1-a}=0. \tag{16} \]

The plus sign in the coefficient of \(d_1\) is taken when \(x_1<x\), and the minus sign when \(x_1>x\).

One may also determine

\[ c_2=z_2^3=\left(\frac{4}{7}a\right)^3, \tag{17} \]

then \(x_2=2\sqrt[3]{c_2}\), and, finally, the correction \(d_2\) from the equation:

\[ d_2^2 \pm x_2 \frac{(3x_2-2a)}{3x_2-a}\, d_2 -\frac{c-c_2}{3x_2-a}=0. \tag{18} \]

The plus sign in the coefficient of \(d_2\) is taken in the case \(x_2<x\), and the minus sign when \(x_2>x\); \(x_2>x\) occurs when \(c_2>c\).

Example. Find the positive root of the equation \(x^3-11x^2-144=0\).

Solution. Without giving the detailed computations, we shall only indicate the values of \(x\) obtained in solving this equation by the first and second variants.

1)

\[ z_1=\sqrt[3]{144}=5.2415;\quad x_1=2z=10.483; \]

\[ a_1=\frac{7}{4}\cdot 5.2415=9.173; \]

\[ a_1<a,\ \text{therefore } x_1<x; \]

\[ d_1\ \text{(from equation 16)}=1.524; \]

\[ x=10.483+1.524=12.007. \]

2)

\[ z_2=\frac{4}{7a}=\frac{44}{7}=6.2855;\quad x_2=2z_2=12.571; \]

\[ c_2=\left(\frac{4\cdot 11}{7}\right)^3=242.8; \]

\[ c_2>c,\ \text{therefore } x_2>x; \]

\[ d_2=0.573\ \text{(from equation 18) and } x=12.571-0.573=11.998. \]

Thus, \(x\) lies within the limits \(12.007 \div 11.998\). It is not difficult to see that the true value of \(x\) will be 12. The greatest error is equal to 0.007, or 0.06%.

Let us solve this same equation, replacing \(x\) in it by \(y+\dfrac{a}{3}\). In the general case, the equation \(x^3-ax^2-c=0\), after transformation, will take the form:

\[ y^3-\left(\frac{a^2}{3}\right)y-2\left(\frac{a}{3}\right)^3-c=0. \]

As applied to the present example, we shall have:

\[ y^3-40.333\,y-242.6=0. \]

Its solution gives:

\[ z_1=\sqrt[3]{242.6}=6.2367;\quad b_1=3.5\sqrt[3]{242.6^3}=136.14; \]

\[ y_1=2z_1=12.4734. \]

The correction \(d_1\) (taking into account that \(y_1>y\), since \(b_1>b\)) is

\[ =-4.990;\quad y=12.4734-4.990=7.463;\quad x=7.463+3.666=11.129. \]

The error is equal to \(12.000-11.129=0.871\), or 7.26%, i.e. many times greater than in the case of the direct solution of the equation

\[ x^3-11x^2-144=0 \]

without reducing it to the form \(x^3-bx-c=0\).

The increase of the error in the case of passing to the transformed equation will occur when \(c\) is considerably greater than \(a\), since under these conditions the discrepancy between the coefficients \(b\) and \(b_1\) in the transformed equation becomes so significant that the quantity \(d_1^3\) can no longer be neglected.

In what follows (§ 9), a method will be considered that makes it possible to substantially reduce the magnitude of the error also in the case of the transformed equation.

§ 6. Approximate solution of an equation of the form

\[ x^3-ax^2+c=0. \]

For equations of this form, the value of the coefficient \(a_1\), for which the condition \(x_1=2\sqrt[3]{c}\) will be satisfied, is found as follows:

\[ a_1=\frac{9}{4}\,z=2.25\sqrt[3]{c}. \tag{19} \]

The value of the constant term \(c_2\), for which, with coefficient \(a\), the condition \(x_2 = 2\sqrt[3]{c_2}\) will be satisfied, is determined by the relation:

\[ c_2=\left(\frac{4}{9}a\right)^3 . \tag{20} \]

In general, this equation is solved analogously to the preceding one, and the corrections \(d_1\) and \(d_2\) are determined from the same equations as in the preceding case. Only in the equation for \(d_2\) must the sign before the constant term be changed to the opposite.

Example. Find the positive root of the equation \(x^3 - 13x^2 + 242 = 0\).

Solution (first variant). \(z_1=\sqrt[3]{242}=6,2317;\quad x_1=2z_1=12,4634;\quad a_1=\)

\[ =\frac{9z_1}{4}=14,021;\quad a_1>a \text{— hence } x_1>x;\quad d_1=1,510;\quad x=12,463-1,510= \]

\[ =10,953. \]

Second variant: \(z_2=\dfrac{4a}{9}=\dfrac{4\cdot 13}{9}=5,7777;\quad x_2=2z_2=11,555;\)

\[ c_2=\left(\frac{4a}{9}\right)^3=192,9;\quad c_2<c \text{— hence } x_2>x; \]

\[ d_2=0,558;\quad x=11,555-0,558=10,997. \]

The true value is \(x=11,0\). Thus, the error in determining \(x\) lies within the limits \(0,047 \div 0,003\), or in percentages \(0,43\text{—}0,03\%\). The error can be reduced further if either of the found values of \(x\) is refined by using the Newton–Raphson rule.

Let us determine \(x\) by reducing the given equation to the form \(x^3-bx+c=0\).

Let \(x=y+\dfrac{a}{3}=y+\dfrac{13}{3}\).

The given equation, after transformation, will take the form:

\[ y^3-\frac{a^2}{3}y-2\left(\frac{a}{3}\right)^3+c=0, \]

or, as applied to the example under consideration,

\[ y^3-\frac{169}{3}y+79,26=0. \]

We solve it:

\[ z_1=\sqrt[3]{79,26}=4,2956;\quad y_1=2z_1=8,5912;\quad b_1=4,5z_1^2=83,032. \]

\[ b_1>\frac{169}{3} \text{— hence } y_1>y. \]

The correction \(d_1=2,0397\) is found from the equation

\[ d_1^2-\left(8,5912-\frac{56,3333}{3\cdot 8,5912}\right)d_1+\frac{83,032-56,333}{3}=0. \]

Then

\[ \begin{aligned} y&=8,5912-2,0397=6,5515,\\ x&=6,5515+4,3333=10,8848. \end{aligned} \]

The true value of \(x\), as was already indicated above, is equal to 11. The error is

\[ 11,0000-10,8848=0,1152, \]

or about \(1,05\%\). Thus, in this case as well, solving the equation reduced to the form \(x^3-bx+c=0\) gives less accuracy than the direct solution of the equation

\[ x^3-ax^2+c=0. \]

§ 7. Approximate solution of an equation of the form

\[ x^3+ax^2-c=0. \]

Denoting \(x\) by \(y-\dfrac{a}{3}\), we obtain the transformed equation in the form

\[ y^3-\frac{a^2}{3}y+2\left(\frac{a}{3}\right)^3-c=0, \]

or

\[ y^3-by+c_1=0, \]

where

\[ b=\frac{a^2}{3}, \qquad c_1=2\left(\frac{a}{3}\right)^3-c. \]

Then we solve the transformed equation by one of the methods considered above. But another way of solving it is also possible.

Since in the given equation \(x^3\) and \(ax^2\) have the same signs, we have

\[ x<\sqrt[3]{c}. \]

Let us take \(x=x_1=0.8\sqrt[3]{c}\,{}^{1}\). Obviously, under these conditions the coefficient of \(x^2\) will no longer be equal to \(a\). The value of the coefficient \(a_1\), satisfying the equation under consideration for \(x_1=0.8\sqrt[3]{c}\), is determined from the equation:

\[ 0.8^3 z^3+a_1\cdot 0.8^2 z^2-z^3=0 \tag{21} \]

\[ a_1=\frac{0.488}{0.640}z, \tag{22} \]

where \(z=\sqrt[3]{c}\).

If \(a_1<a\), then \(x_1\) is greater than the true value; in the opposite case, conversely.

The magnitude of the correction \(d_1\) is determined by jointly solving the equations:

\[ (x_1+d_1)^3+(x_1+d_1)^2a-c=0 \]

\[ x_1^3+a_1x_1^2-c=0, \]

or

\[ (a+3x_1)d_1^2+2(ax_1+3x_1^2)d_1+x_1^2(a-a_1)=0 \tag{23} \]

The plus sign before the coefficient of \(d_1\) is taken when \(a_1>a\), and the minus sign when \(a_1<a\).

Example. Find the positive root of the equation \(x^3+10x^2-375=0\).

Solution. Let us take

\[ x_1^3=0.8\sqrt[3]{375}=0.8\cdot 7.2112=5.76896, \]

or, rounded,

\[ x_1=5.769. \]

\[ a_1=\frac{488z}{640}=\frac{61}{80}\sqrt[3]{375}=\frac{61\cdot 7.2112}{80}=5.498. \]

Since \(a_1<a\), \(x_1\) has been obtained with excess. We find the correction \(d_1\) from equation (23), represented in the form:

\[ \left(\frac{a}{x_1}+3\right)d_1^2-(2a+3x_1)d_1+x_1(a-a_1)=0 \]

After substituting the values of \(a\), \(x_1\), and \(a_1\), we obtain:

\[ \left(\frac{10}{5.769}+3\right)d_1^2-(2\cdot 10+3\cdot 5.769)d_1+5.769(10-5.498)=0 \]

\[ 4.733d_1^2-37.307d_1+25.972=0 \]

\[ d_1^2-7.882d_1+5.4874=0, \]

\(^{1}\) In what follows, the general case will be considered, allowing \(x\) to be expressed as \(m\sqrt[3]{c}\), where \(m\ne 1\).

whence

\[ d_1=3.941-\sqrt{15.5248-5.4874} \]

\[ =3.941-3.168=0.773. \]

Then \(x=x_1-d_1=5.769-0.773=4.996\).

The true value is \(x=5\). The error is \(0.004\), or \(0.1\%\), i.e., the accuracy is quite sufficient for technical calculations. It should be noted that for equations of this type the substitution of \(x\) by \(y-\dfrac{a}{3}\) leads to a large error in determining \(y\), and hence also \(x\).

§ 8. Approximate solution of an equation of the form

\[ x^3+ax^2+c=0 \]

Equations of this kind have no positive roots. Changing the sign of the odd power to the opposite one, we obtain the equation \(x^3-ax^2-c=0\), already considered earlier. The positive root found in solving this equation must be taken with the opposite sign.

Example. Find the root of the equation \(x^3+2.5x^2+4.5=0\).

Solution. We represent the given equation in the form: \(x^3-2.5x^2-4.5=0\).

Take

\[ x_1=2\sqrt[3]{4.5}=2\cdot1.651=3.302. \]

According to the preceding (§5),

\[ a_1=\frac{7z}{4}=\frac{7\cdot1.651}{4}=2.889. \]

Since \(a_1>a\), \(x_1\) is greater than the true value \(x\).

We find the correction \(d_1\) from expression (16):

\[ d_1^2-x_1\frac{(3x_1-2a)}{3x_1-a}d_1+\frac{(a_1-a)x_1^2}{3x_1-a}=0, \]

\[ d_1^2-3.302\frac{(3\cdot3.302-2\cdot2.5)}{3\cdot3.302-2.5}d_1+ \frac{(2.889-2.5)3.302^2}{3\cdot3.302-2.5}=0, \]

or

\[ d_1^2-2.1874d_1+0.5727=0, \]

whence

\[ d_1=1.0937-\sqrt{1.1962-0.5727}, \]

or

\[ d_1=1.0937-0.7896=0.3041. \]

Then

\[ x\simeq 3.302-0.304=2.998. \]

The true value is \(x=3\). The error is \(0.002\), or \(0.067\%\).

§ 9. Determination of the value of \(x\) without solving equations for the magnitude of the correction

In the preceding paragraphs, when solving equations of the third degree of various types, the value of \(x\) was taken equal either to \(2\sqrt[3]{c}\) or to \((0.8\div0.9)\sqrt[3]{c}\), and was subsequently refined by means of a correction. In doing so, the possibility is by no means excluded that between the adopted and the true value of \(x\) there may be so considerable a discrepancy that even the correction will not remedy the situation; all the more so because, for values \(d>1\), the accuracy of the correction itself becomes doubtful, since in this case neglecting the quantity \(d^3\) gives a significant error in determining \(d\). Therefore it is more rational to take \(x\) equal not to \(2\sqrt[3]{c}\), but

\(m\sqrt[3]{c}\), although such an assumption will ultimately require the solution of a cubic equation with respect to \(m\). Indeed, suppose, for example, that we have the equation \(x^3-bx-c=0\). Substitute \(m\sqrt[3]{c}\) for \(x\) in it and divide all terms by \(c\). We obtain:

\[ \frac{m^3c}{c}-\frac{bm\sqrt[3]{c}}{c}-\frac{c}{c}=0, \]

or

\[ m^3-b_1m-1=0, \tag{24} \]

where

\[ b_1=\frac{b}{\sqrt[3]{c^2}}. \]

The solution of equation (24) is simpler than that of the original one from which it was obtained.

Indeed, writing equation (24) in the form \(m^2=b_1+\dfrac{1}{m}\), or

\[ m=\left(b_1+\frac{1}{m}\right)^{1/2} \tag{25} \]

and considering the right-hand side of expression (25) as a binomial, we obtain, limiting ourselves to two terms of the binomial,

\[ m=b_1^{1/2}+\frac{1}{2}\frac{1}{m}b_1^{-1/2} =\sqrt{b_1}+\frac{1}{2m\sqrt{b_1}}, \]

or

\[ m^2-m\sqrt{b_1}-\frac{1}{2\sqrt{b_1}}=0. \tag{26} \]

Thus, the approximate solution of the cubic equation (24) is reduced to solving the quadratic equation (26)

\[ m=\frac{\sqrt{b_1}}{2}\pm\sqrt{\frac{b_1}{4}+\frac{1}{2\sqrt{b_1}}} \tag{27} \]

or

\[ m=\frac{b_1\pm\sqrt{b_1^2+2\sqrt{b_1}}}{2\sqrt{b_1}}. \tag{27a} \]

It is not difficult to see that the sign before the radical in the numerator of expression (27a) must be \(+\), since \(m\) cannot be a negative number.

Example. Let us solve by this method the equation

\[ x^3-23,4x-8=0, \]

already solved in § 1.

The transformed equation will take the form:

\[ m^3-\frac{23,4m}{\sqrt[3]{8^2}}-1=0, \quad \text{or} \quad m^3-5,85m-1=0. \]

In this equation \(b_1=5,85\),

\[ m=\frac{5,85+\sqrt{5,85^2+2\sqrt{5,85}}}{2\sqrt{5,85}} =\frac{5,85+\sqrt{34,2225+2\cdot 2,4187}}{2\cdot 2,42} \]

or

\[ m=\frac{5,85+\sqrt{39,0599}}{4,84} =\frac{5,85+6,250}{4,84} =\frac{12,10}{4,84}=2,5. \]

Since \(x=m\sqrt[3]{c}\), in the present case \(x=2,5\sqrt[3]{8}=5,0\), which is the true root of the equation under consideration. In § 1 the value \(5,02\) was obtained for \(x\), i.e., with an error of \(0,4\%\).

It should be noted, however, that the degree of accuracy in determining the quantity \(m\) from expressions (27) and (27a) depends to a considerable extent on the absolute magnitude of \(b_1\), as can be seen from Table 1.

Table 1

True values of \(m\) and values calculated from expression (27), as functions of \(b_1\)

Values of \(b_1\)	Values of \(m\), true	Values of \(m\), calculated from expression (27)	\(\Delta m\)	\(\Delta m\) in %
0,309	1,1	1,2660	0,1660	15,10
0,607	1,2	1,2804	0,0804	6,70
0,921	1,3	1,347	0,047	3,62
1,246	1,4	1,4296	0,0296	2,12
1,583	1,5	1,5186	0,0186	1,24
1,935	1,6	1,6136	0,0136	0,85
2,302	1,7	1,7099	0,0099	0,58
2,684	1,8	1,8072	0,0072	0,40
3,084	1,9	1,9057	0,0057	0,30
3,500	2,0	2,0042	0,0042	0,21
3,934	2,1	2,1036	0,0036	0,17
4,385	2,2	2,2029	0,0029	0,13
5,343	2,4	2,4012	0,0012	0,05

As \(b_1\) increases, the value of \(m\) also increases, and at the same time the discrepancy between the true and calculated values of \(m\) decreases. Indeed, from the expression \(m=\dfrac{\sqrt{b_1}}{2}+\sqrt{\dfrac{b_1}{4}+\dfrac{1}{2\sqrt{b_1}}}\) it follows that, as \(b_1\) increases, its influence on the magnitude of the radicand will decrease, and \(m\) will approach \(\sqrt{b_1}\). This, incidentally, is also clear from the equation \(m^3-b_1m-1=0\): for sufficiently large \(b_1\), \(m\cong\sqrt{b_1}\). Thus, for \(b_1=100\), the difference between the true and approximate values of \(m\) will be only \(\dfrac{1}{290}\) (or \(0,00345\)), i.e., only \(0,0345\%\). In other words, for \(b_1>2,5\), expression (27) gives results quite sufficient for practical purposes.

For an equation of the form \(x^3-bx+c=0\), the value of \(m\) is obtained from an expression analogous to expression (27), namely:

\[ m=\frac{b_1}{2}+\sqrt{\frac{b_1}{4}-\frac{1}{2\sqrt{b_1}}}, \tag{28} \]

where

\[ b_1=\frac{b}{\sqrt[3]{c^2}}. \]

Table 2 gives the true values of \(m\) and those calculated by formula (28), as well as the absolute error and the error in percent, i.e., \(\dfrac{100\Delta m}{m\ \text{(true)}}\).

Table 2

True values of \(m\) and those computed by formula (28), as a function of \(b_1\)

Values of \(b_1\)	Values of \(m\), true	Values of \(m\), computed from expression (28)	\(\Delta m\)	\(\Delta m\) in %
2.1192	1.1	1.1646	0.0646	+5.87
2.733	1.2	1.2537	0.0537	4.48
2.4592	1.3	1.3282	0.0282	2.19
2.6743	1.4	1.4192	0.0192	1.37
2.9133	1.5	1.5135	0.0135	0.90
3.1850	1.6	1.6110	0.0110	0.69
3.4782	1.7	1.7083	0.0083	0.50
3.7944	1.8	1.8055	0.0055	0.31
4.1363	1.9	1.9047	0.0047	0.24
4.5000	2.0	2.0036	0.0036	0.18
5.2945	2.2	2.2031	0.0031	0.14
6.1766	2.4	2.4025	0.0025	0.10
6.6500	2.5	2.5013	0.0013	0.052

Having determined \(b_1\) from the expression

\[ b_1=\frac{b}{\sqrt[3]{c^2}}, \]

one can, using Tables 1 and 2 (depending on the type of equation), find the nearest value of \(m\) corresponding to the value of \(b_1\) found. A more exact value of \(m\) can be found as follows: after solving equation (27) or (28), one finds from Tables 1 or 2 the limits between which \(m\) lies. If, for example, for the equation \(m^3-b_1m+1=0\) one obtained \(m=1.55\), then from Table 2 it is seen that the possible error lies within the limits \(0.9 \div 0.69\%\). In the present case it will be of the order of \(0.78\%\). Dividing the value found for \(m\) by

\[ 1+\frac{\Delta m}{100}, \]

we obtain a value differing from the true one by hundredths of a percent, which is more than sufficient in solving technical problems. For example, in determining the composition of an electric train, or the permissible heating of traction motors, such accuracy is even excessive, since an error of \(1\%\) is quite admissible here.

For solving an equation of the form \(m^3+b_1m-1=0\), equations (27) or (28) cannot be used, since the expression

\[ m^2=\left(\frac{1}{m}-b_1\right) \]

or

\[ m=\left(\frac{1}{m}-b_1\right)^{1/2} \]

ultimately nevertheless leads to an equation of the third degree. Therefore \(m\) must be determined approximately, using Table 3.

Table 3

Value of \(m\) as a function of \(b_1\)

\(m\)	\(b_1\)	\(m\)	\(b_1\)
0.05	19.9975	0.55	1.5157
0.075	13.3244	0.60	1.3066
0.10	9.9900	0.65	1.1160
0.15	6.6441	0.70	0.9386
0.20	4.9600	0.75	0.7708
0.25	3.9375	0.80	0.6100
0.30	3.2433	0.85	0.4565
0.35	2.7346	0.90	0.3011
0.40	2.3400	0.95	0.1501
0.45	2.0197	1.00	0.0000
0.50	1.7500

If, for example, \(b_1\), found from the equation \(b_1=\dfrac{b}{\sqrt[3]{c^2}}\), is equal to 3, then \(m\) may be taken equal to

\[ 0.35-\frac{(3.0000-2.7346)0.05}{3.2433-2.7346}=0.35-0.026=0.324. \]

§ 10. Solution of complete equations of the third degree

The solution of a complete equation of the third degree can be reduced to three stages: 1) transformation of the complete equation into an incomplete one by replacing \(x\) by \(y \pm \dfrac{a}{3}\); 2) transformation of the resulting incomplete equation into an equation of the form \(m^3-b_1m-1=0\), \(m^3-b_1m+1=0\), or \(m^3+b_1m-1=0\); 3) solution of the resulting equation with respect to \(m\) and determination of

\[ y=m\sqrt[3]{c_1}\quad\text{and}\quad x=y\pm\frac{a}{3}. \]

Example. Find the positive root of the equation

\[ x^3-6x^2+10x-8=0. \]

If we denote \(x\) by \(y+2\), then the given equation can be represented in the form \(y^3-2y-4=0\), i.e., in the form of an equation conditionally assigned to type 1. Denoting \(y=m\sqrt[3]{4}\) and substituting this value into the resulting equation, we shall have:

\[ m^3-0.7937\,m-1=0,\quad \text{where } 0.7937=b_1. \]

Subsequently the resulting equation may be solved either by formula (27), or, using Table 1, one may find \(m\) corresponding to the found value \(b_1\). In view of the small value of \(b_1\), the possible magnitude of the error when using equation (27) will be very significant, lying within the limits \(6.7 \div 3.62\%\). Therefore it is better to find \(m\) from Table 1 by interpolation, although in this case also an error is inevitable, but one considerably smaller than when determining \(m\) from equation (27).

For the example under consideration,

\[ m=1.2+\frac{(0.7937-0.6070)0.1}{0.921-0.607}=1.2+0.059\cong 1.26. \]

Then \(y=1.26\sqrt[3]{4}=2.0001\). We take \(y=2\). Then \(x=y+2=4\). Verification of the obtained value of \(x\) by the Newton–Raphson formula gives for \(f(x)\) the value 0, which means that the correction is equal to zero and the value has been obtained with absolute accuracy. Complete equations of the third degree reducible to the form \(y^3-by+c=0\) or \(y^3+by-c=0\) are solved analogously. In the first case \(m\) is determined either from equation (28) or from Table 2; in the second case—from Table 3, with subsequent verification by the Newton–Raphson formula.

Brief conclusions

1. Incomplete equations of the third degree may be conditionally divided into 8 groups, which ultimately may be reduced to 3 principal groups.

1. The solution of equations of the first two groups reduces to determining an approximate value of the unknown, conventionally taken to be equal to the doubled cube root of the constant term, and to finding a correction determined from the equations.

2. In solving equations of the 3rd group, the assumed value of the unknown is taken equal to \(0.7 \div 0.9\) cube roots of the constant term.

The value of the coefficient before \(\sqrt[3]{c}\) is chosen so that the difference between the coefficients \(b\) and \(b_1\) will be minimal.

1. Incomplete cubic equations containing \(x^2\) may be solved either by a method analogous to the method proposed for solving the first three groups of equations, or reduced to one of these groups by replacing \(x\) by \(y \pm \dfrac{a}{3}\).

In the first case the determination of the magnitude of the correction \(d\) is more complicated, but the accuracy is higher.

1. Instead of the assumed value \(x = 2\sqrt[3]{c}\), it is more expedient to take

\[ x = m\sqrt[3]{c}. \]

The value of \(m\) for the first two groups of equations is determined from the expressions

\[ m = \frac{\sqrt{b_1}}{2} + \sqrt{\frac{b_1}{4} + \frac{1}{2\sqrt{b_1}}} \quad \text{or} \quad m = \frac{\sqrt{b_1}}{2} + \sqrt{\frac{b_1}{4} - \frac{1}{2\sqrt{b_1}}}, \]

where, if \(b > 3.5\), the error will not exceed \(0.5\%\). Otherwise, the found value of \(m\) must be refined either by means of the Newton–Raphson formula or by means of Tables 1 and 2.

For equations of the third group, the value of \(m\) is determined from Table 3 by interpolation, in accordance with the coefficient \(b_1\).

Equations containing \(x^2\) are first reduced to equations of the first three groups, for which the determination of the value \(m\) is indicated above.

1. The two proposed methods for approximate solutions of incomplete and complete equations of the third degree are distinguished by their simplicity, give an accuracy quite sufficient for technical calculations, and do not require cumbersome computations.

1. E. Whittaker and R. Robinson, The Mathematical Processing of Observational Results, ONTI, 1935, pp. 77–78. ↩