 # Mark Kac’s First Publication: A Translation of "O nowym sposobie rozwiązywania równań stopnia trzeciego" – Suggestions for Use by Teachers and Students

Author(s):
David Derbes (University of Chicago Laboratory Schools, retired)

Even world-class mathematicians start from modest beginnings, and were once high school students themselves, perhaps even among those made uncomfortable when approached by the school’s principal. Here is a calculation well within the grasp of most algebra students, and yet it was good enough to spark a major career. It is hoped that some young people will find this an inspiring story and Mark Kac an appealing person. Maybe there are other old problems out there for which they can find a new solution all their own, or even a first solution. Mathematics is a deep subject with a rich history. It invites exploration.

The cubic itself is a fascinating topic. An argument can be made for its being the springboard for the whole of today’s algebra. Its historical and cultural importance is well known. For example, Feynman wrote to his family while touring Greece that he was disheartened by the locals’ excessive respect for the great accomplishments of the past:

They were very upset when I said that the thing of greatest importance to mathematics in Europe was the discovery by Tartaglia that you can solve a cubic equation—which, altho [sic] it is very little used, must have been psychologically wonderful because it showed that a modern man could do something no ancient Greek could do, and therefore helped in the renaissance which was the freeing of man from the intimidation of the ancients—what they are learning in school is to be intimidated into thinking they have fallen so far below their super ancestors [Feynman 2006, 327].

Considering how simply the Cardano formula can be derived, as a consequence of the quadratic—as shown by Viète—it’s surprising that it is not often taught in the schools. Here’s how that might be done.

First, it should be pointed out that the general cubic equation,

$Ax^{3} + Bx^{2} + Cx + D = 0$

can be reduced to the monic

$x^{3} + bx^{2} + cx + d = 0$

simply by dividing by $A$. Then, use the Tschirnhaus18 transformation [Tignol 2001, 68]: let

$x = y - \frac{1}{3}b$

and, substituting this expression for $x$ into the monic polynomial, rewrite the equation in terms of $y$. This removes the quadratic term (the same trick can be used for any polynomial, e.g., let $x=y-\frac{1}{2}b$ for a quadratic). One returns to the standard form (∗) considered by Kac,

$y^3 + py + q = 0,$

where $p = c - \frac{1}{3}b$, $q = d - \frac{1}{3}bc + \frac{2}{27}b^{3}$. Thus any cubic can be turned into the standard form for which the Cardano formula holds. Now comes the master stroke given in [Viète 1615, 287]: let

$y = u - \frac{p}{3u}$

Plugging this into the previous equation results in the wonderfully simple

$u^3 + q - \frac{p^3}{27u^3} = 0$

Multiply both sides by $u^3$, and obtain a quadratic in $u^3$:

$u^6 + qu^3 – \frac{p^3}{27} = 0$

for which the roots are given by the familiar quadratic formula,

$u^{3} = \frac{-q \pm \sqrt{q^{2} + \frac{4p^{3}}{27}}}{2} = -\frac{q}{2} \pm \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3}\right)^{3}}$

Let $u_1$ correspond to the real cube root of the quantity with the plus sign, and $u_2$ to that of the minus. That means

$u^{6} + qu^{3} - \frac{p^{3}}{27} = (u^{3} - u_{1}^{3})(u^{3} - u_{2}^{3})$

so that $u_{1}^{3}u_{2}^{3} = -(p/3)^{3}$, or $u_{2} = - (p/3u_{1})$. Finally a root of the cubic is given by

$y = u_{1} - \frac{p}{3u_{1}} = u_{1} + u_{2} = \sqrt{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3}\right)^{3}}} + \sqrt{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} +\left(\frac{p}{3}\right)^{3}}},$

exactly the Cardano formula.19

Once a student has learned to solve the cubic, Kac’s method should be presented. Many students are presented with only one way to do something; it’s worthwhile to show two or three ways to reach a result. For students in calculus, a cute exercise is to show that a function of the form (∗) will have three distinct real roots only if the product of its local maximum and its local minimum is negative; only then will the graph cross the axis three times. And of course this is just the condition that the expression (8) is negative.

It’s true that cubic equations do not arise in the sciences as often as quadratics, but they do crop up occasionally, for example in chemistry problems involving a particular type of equilibrium reaction, or in classical mechanical problems involving rotations and eigenvectors of 3 × 3 matrices. Even so, the cubic should be taught, if only to highlight the creativity of great mathematicians who recast these equations into forms more easily solved. Beyond the inventive algebra, the dramatic human story behind the cubic has few peers in mathematical history for intrigue and conflict.

Finally, should a student become interested in the theory of equations generally, she might begin by retracing Ferrari’s path to the quartic, and perhaps even look at Abel’s proof (1824) that the general fifth-degree equation cannot be solved with radicals. Resources suitable for talented and motivated high school students include [Livio 2005], [Pesic 2003], and [Alekseev 2004]. That might in turn lead her to Galois theory and its creator, Évariste Galois (1811–1832), whose tragic biography is perhaps the most romantic story in mathematics. In addition to [Livio 2005], [van der Waerden 1970], [Bewersdorff 2006], and [Tignol 2001], works on Galois theory that can be recommended highly to beginners include [Stewart 2015] and [Artin 2007]. And to come full circle, a beautiful introduction to the ideas of Galois theory is given by Kac and Ulam in [Kac and Ulam 1968, 56–60].

##### Notes

 Ehrenfried W. von Tschirnhaus (1651–1708), mathematician and possible inventor of European porcelain.

 It should be noted that the cubic has two other roots, but they're easily found. There are three cube roots of $u_{1}^{3}$: $u_{1}$, $\omega u_{1}$, and $\omega^{2}u_{1}$, where $\omega = \exp(\frac{2}{3}\pi i)$ is a primitive cube root of 1, and $\frac{1}{\omega }= \omega^{2}$. Keeping the conditions $u_{1}u_{2} = - p/3$ and $y = u - \frac{p}{3u}$, but replacing $u_{1}$ first by $\omega u_{1}$ and then by $\omega^{2}u_{1}$ gives $y_{2} = \omega^{2}u_{1} + \omega u_{2}$ and  $y_{3} = \omega u_{1} + \omega^{2}u_{2}$ respectively. Simple!

David Derbes (University of Chicago Laboratory Schools, retired), "Mark Kac’s First Publication: A Translation of "O nowym sposobie rozwiązywania równań stopnia trzeciego" – Suggestions for Use by Teachers and Students," Convergence (April 2021)

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