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Mark Kac’s First Publication: A Translation of "O nowym sposobie rozwiązywania równań stopnia trzeciego" – The English Translation

David Derbes (University of Chicago Laboratory Schools, retired)

What follows is my translation of Mark Kac’s article11 from Polish to English. Though ignorant of Polish, I have some Russian; that, with the help of Google Translate, was enough for a reasonable rendering.12 Note that there is one alteration from the original. A third handwritten notation in Kac’s copy crosses out “(4)” cited in the last sentence of Kac’s work, before the Editor’s Note, and substitutes “(6)”. This substitution is incorporated in the English translation and in the transcribed Polish version of Kac's original article, "O nowym sposobie rozwiązywania równań stopnia trzeciego'' (pdf).


On a new way of solving equations of the third degree

To solve the equation \(z^{3} + az^{2} + bz + c = 0\), we substitute

\[z = x - \frac{1}{3}a\]

and we obtain an equation without the unknown in the second power:

\[\hspace{3in}x^{3} + px + q = 0, \hspace{2.1in}(1)\]

where \(p\) and \(q\) are terms dependent on \(a\), \(b\), and \(c\). Here I want to give—as it seems to me—a new way of solving equations of type (1).

My way is to find certain numbers \(A\), \(B\), \(m\), \(n\), such that for every value of \(x\) this equality holds:

\[\hspace{2in}x^{3} + px + q = A(x + m)^{3} - B(x + n)^{3}. \hspace{1.45in}(2)\]

Then we get the equation in the form

\[\hspace{2.5in}A(x + m)^{3} - B(x + n)^{3} = 0 \hspace{1.8in}(3)\]

and it will be easy to solve this; for we observe that the left side of the equation breaks down into factors:

\(A(x + m)^{3} - B(x + n)^{3} \)
\(\hspace{.1in}= \left[\sqrt[3]{\!A}(x + m) - \sqrt[3]{B}(x + n)\right]\cdot\left[\sqrt[3]{\!A^{2}}(x + m)^{2} + \sqrt[3]{\!AB}(x + m)(x + n) + \sqrt[3]{B^{2}}(x + n)\right]\!.\)

Setting this product equal to zero, we have two possibilities:

\[\hspace{1.8in}\begin{array}{c}\hspace{.8in}\sqrt[3]{A}(x + m) - \sqrt[3]{B}(x + n) = 0, \hspace{1.4in}(4)\\ \sqrt[3]{\!A^{2}}(x + m)^{2} + \sqrt[3]{\!AB}(x + m)(x + n) + \sqrt[3]{B^{2}}(x + n) = 0.\hspace{.5in}(5)\end{array}\]

As we can see from equation (4), it will be easy to find one of the roots of the given equation (1).

So let's find the numbers \(A\), \(B\), \(m\), and \(n\). Expanding the right side of the equality (2), we obtain:

\[x^{3} + px + q = (A - B)x^{3} + 3(Am - Bn)x^{2} + 3(Am^{2} - Bn^{2})x + (Am^{3} - Bn^{3}).\]

If two polynomials of one variable have equal values for every value of this variable, then the coefficients of correspondingly equal powers of the variable must be equal:

\[\begin{array}{c}A - B = 1,\\ Am - Bn = 0,\\ Am^{2} - Bn^{2} = \frac{1}{3}p,\\ Am^{3} - Bn^{3} = q.\end{array}\]

 We are to solve the system of 4 equations in 4 unknowns. From the first equation we find \(A = B + 1\) and substitute into the remaining equations:

\[\begin{array}{c}B(m - n) + m = 0,\\ B(m^{2} - n^{2}) + m^{2} = \frac{1}{3}p,\\ B(m^{3} - n^{3}) + m^{3} = q.\end{array}\]

Now we find from the group \(B(m-n) = -m\) and substitute into the two remaining equations:

\[\begin{array}{c}-m(m + n) + m^{2} = \frac{1}{3}p,\\ -m(m^{2} + mn + n^{2}) + m^{3} = q.\end{array}\]

After the simplifications we get:

\[\begin{array}{c}mn = -\frac{1}{3}p,\\ m + n = \frac{3q}{p},\end{array}\]

we assume that \(p \neq 0\).  (The case when \(p = 0\) gives the equation \(x^{3} + q = 0\), whose solution we will not find difficult.)

We find the values of \(m\) and \(n\) as the roots of the quadratic equation

\[\hspace{2.5in}u^{2} - \frac{3q}{p}u - \frac{1}{3}p = 0. \hspace{2.2in}(6)\]

If it turns out that this equation has its discriminant positive, then it has two roots which are not equal. By assuming one of these values for \(m\), and the other for \(n\), we can easily calculate the values of \(A\) and \(B\), namely

\[B = \frac{-m}{m-n}, \quad A = \frac{- n}{m-n}.\]

Substituting these values into (4) and multiplying both sides of this equation by \(\sqrt[3]{m-n}\), we obtain the equation

\[- \sqrt[3]{n}(x + m) + \sqrt[3]{m}(x + n) = 0.\]

Hence, after rearranging, we have the following equation:

\[\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)x = m\sqrt[3]{n} - n\sqrt[3]{m}.\]

By transforming the right side, we will have

\[\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)x = \sqrt[3]{mn}\left(\!\sqrt[3]{m} - \sqrt[3]{n}\,\right)\left(\!\sqrt[3]{m} + \sqrt[3]{n}\,\right),\]

so finally we will have

\[\hspace{2.1in}x = \sqrt[3]{mn}\left(\!\sqrt[3]{m} + \sqrt[3]{n}\,\right). \hspace{2.2in}(7)\]

Substituting the values of \(m\) and \(n\), determined from equation (6), we will obtain the formula known as Cardano's formula.


Młody Matematyk Editor's Note.

Granting room to the method presented by Mr. Marek Katz, who is a student of the 8th grade of the Krzemieniec Lyceum preparatory school, the Editorial Board must add a few comments.

The author of the article limited himself to considering the case when the discriminant of  equation (6) is positive, i.e., the case when the expression

\[\hspace{2.2in}\left(\frac{1}{2}q\right)^{2} + \left(\frac{1}{3}p\right)^{3} \hspace{2.3in}(8)\]

has a positive value.

From the theory of the third degree equation, it is known that in this case equation (1) has one real root. The method described in the article relates to finding this root.

In the case when expression (8) is zero, the discriminant of equation (6) is zero, so this equation has a double root:

\[m = n = \frac{3q}{2p}.\]

But in this case, formula (7) takes the form \(x = \frac{3q}{p}\), and it is not difficult to verify by direct substitution that this formula gives the root of equation (1).

In the event that expression (8) has a negative value, \(m\) and \(n\) are not real numbers. The case here is known as the casus irreducibilis—peculiar in that in this case the third degree equation has three real roots, but they cannot be determined via an algebraic path. Clearly, the particular method described by Mr. M. Katz, fails in this case. Indeed, the equation

\[(x - 1)(x - 2)(x + 3) = 0\]

has three roots: \(+1,\) \(+2\), \(-3\). Expanding the left side, we get the equation

\[x^{3} - 7x + 6 = 0,\]

so we have these values: \(p = -7\), \(q = 6\). Equation (6) takes the form13

\[u^{2} + \frac{18}{7}u + \frac{7}{3} = 0.\]

It is easy to check that this equation does not have real roots, so the real values of \(m\) and \(n\) cannot be determined from it; but it does not follow that the given third-degree equation has no roots.


[11] Originally published as “O nowym sposobie rozwiązywania równań stopnia trzeciego,” Młody Matematyk, Rok 1, Kwiecień-Maj 1931, Nr. 4–5, pp. 69–71 ("On a new way of solving equations of the third degree," Young Mathematician 1(4–5) (April–May 1931): 69–71.)

[12] The translation was checked by Zbigniew Kantorosinski, a scholar at the Library of Congress and a native speaker of Polish. He had not only the translation, but my retyped Polish of the original. Editors’ Note: During the blind refereeing process for this article, the author’s Polish-to-English translation was also vetted by a mathematician fluent in both English and Polish, based on the published version of Kac’s article in Młody Matematyk.

[13] In the original article, the fraction \(\frac{18}{7}\) was erroneously written as \(\frac{6}{7}\).


David Derbes (University of Chicago Laboratory Schools, retired), "Mark Kac’s First Publication: A Translation of "O nowym sposobie rozwiązywania równań stopnia trzeciego" – The English Translation," Convergence (April 2021)