# Mark Kac’s First Publication: A Translation of "O nowym sposobie rozwiązywania równań stopnia trzeciego" – Casus Irreducibilis

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

The comment by Rusiecki, though correct, seems off-target. In the casus irreducibilis, or “irreducible case”, the three real roots are usually given in terms of a trigonometric function (see below); a transcendental expression, not algebraic. The “failure” however lies not with Kac’s derivation, but with Cardano’s formula. Kac set out to find a derivation of Cardano’s formula that did not assume the form of the answer, and he succeeded. Even in the “irreducible case”, Cardano’s formula provides a correct expression for a real root; the purported “failure” is that this is not explicitly done in terms of real numbers alone—roots of complex numbers arise. The argument is made (particularly by Tignol [2001, 16–20] and Nahin [1998, 25]) that, contrary to what legions of high school students are taught, imaginary numbers found a home in algebra not because of quadratics, but because of cubics. But this invites a longer discussion.

In Cardano’s time, not merely the square roots of negative numbers, but negative numbers themselves were viewed as “fictitious”, or, if encountered as the root of a quadratic, “impossible” [Kline 1972, 252–253], not least because most algebraic questions were couched in geometric terms: a geometric problem requiring a negative answer violated common sense. When at last Cardano derived the del Ferro-Tartaglia expressions, he did so via a geometric argument [Cardano 1545, 96–98].14 But when the root of a cubic known to have a real root was expressed by Cardano’s formula in terms of the square roots of negative numbers, there had to be a reckoning. Indeed, Rafael Bombelli (1526–1572)—in the case of the cubic15—and Cardano himself—in the case of a quadratic—were able to show through manipulation of complex quantities that a sum or a product of complex quantities could in fact equal a real number:

If it should be said, Divide $10$ into two parts, the product of which is $30$ or $40$, it is clear that this case is impossible. Nevertheless, we will work thus: We divide $10$ into two equal parts, making each $5$. These we square, making $25$. Subtract $40$, if you will, from the $25$ thus produced,  . . . leaving a remainder of $-15$, the square root of which added to or subtracted from $5$ gives parts the product of which is $40$. These will be $5 + \sqrt{-15}$ and $5 - \sqrt{-15}$. . . . Putting aside the mental tortures involved,16 multiply $5 + \sqrt{-15}$ by $5 - \sqrt{-15}$, making $25 - (-15)$, which is $+40$. Hence this product is $40$. . . . So progresses arithmetic subtlety, the end of which, as is said, is as refined as it is useless [Cardano 1545, 219–220].

It is likely that Rusiecki would have described the extraction of complex roots of a quadratic equation as algebraic. His disqualification of Cardano’s formula as algebraic can only be due to its inclusion, in the irreducible case, of a radical of a number not in the base field $R$ of the equation’s coefficients. Well, is there perhaps some method of writing, in general, the complex expressions occurring in the casus irreducibilis as $a+i\sqrt{b}$, where a and b are real? That would apparently render Cardano’s formula “algebraic”, in accord with the quadratic formula.

Whether the roots of a cubic or any polynomial with real coefficients can be expressed in terms of real roots only (perhaps multiplied by $i$) has been answered. One of the first investigations is due to Mollame [1890], who found that, given the casus irreducibilis, if the coefficients of the polynomial satisfied the relation (10) in his article,

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

then a real root was expressed simply as

$x = \sqrt[3]{2q}.$

For the general irreducible case, however, Mollame was skeptical that real radicals alone could be used.

Van der Waerden [1970, 189–190] stated unequivocally that “it is actually impossible to solve the equation [in the irreducible case] by real radicals unless the equation is reducible in the base field K” (in which the coefficients lie), and provided the proof. More recently, Isaacs [1985] has shown that, given an irreducible polynomial with rational real coefficients, and which has only real roots, then if it has any root expressible in terms of real radicals, the degree of the polynomial must be a power of 2. That clearly rules out the cubic. In short, what Rusiecki found lacking in the Cardano formula is simply not to be found at all, at least in terms of real roots.

But, as Rusiecki implied, the expression of a cubic’s three real roots in the casus irreducibilis as real expressions can be accomplished, by means of a trigonometric function. This was accomplished by François Viète (1540–1603), in a posthumous publication [Viète 1615, 174–175]. In modern terms, the argument is this [Nahin 1998, 22–23]. From (8)’s being negative and the reality of $p$, $q$, it follows that $p < 0$. Then we can say

$p = -3a^{2}, \quad q = - a^{2}b,$

where we can assume $a > 0$ because $a$ appears only as $a^{2}$, and the original cubic becomes

$y^{3} - 3a^{2}y - a^{2}b = 0. \hspace{.2in}(†)$

The irreducible case, in which expression (8) must be negative, leads to

$\frac{|b|}{2a} < 1$

so that we can set, without loss of generality,

$\frac{b}{2a} = \cos \theta$

for some angle $\theta$, $0 < \theta < \pi$. Viète, a master of trigonometry,17 knew the identity

$\cos 3\phi = 4\cos^{3}\phi - 3\cos \phi. \hspace{.2in} (‡)$

This follows easily from the addition and double angle formulas, expanding $\cos 3\phi$ as $\cos (2\phi + \phi)$. Multiplying both sides of (‡) by $2a^{3}$ gives

$(2a\cos \phi)^{3} - 3a^{2}(2a\cos \phi) - a^{2}(2a\cos 3\phi) = 0,$

which is identical to (†), with $y = 2a\cos \phi$, provided the angle $3\phi$ is such that

$\cos 3\phi = \frac{b}{2a} = \sqrt{\frac{(\tfrac{1}{2}q)^{2}}{(-\frac{1}{3}p)^{3}}}.$

Thus, we obtain the three roots of the cubic (†) as

$y_{k} = 2a\cos \phi_{k}, \quad \phi_{k} = \tfrac{1}{3}\cos^{-1}(b/2a) + \tfrac{2}{3}\pi(k-1),\; k = \{1, 2, 3\}.$

Incidentally, the restriction “$m\neq n!$” penciled in by Kac should be taken not as a requirement, but simply as a warning to the reader that these are not necessarily equal. They can be, as Rusiecki pointed out, but in that case, there is a double root.

##### Notes

[14] Very clear discussions of the geometric argument cast into algebraic language are given by Struik [1969, 64–65, see notes 3 and 5 in particular] and Bewersdorff [2006, 5–6]. According to Kline [1972, 253], the first to give both positive and negative roots of a quadratic was Albert Girard (1595–1632) in his L’Invention nouvelle en l’algèbre (1629).

[15] See [Fauvel and Gray 1987, 265].

[16] The Latin original is dismissis incruciationibus, literally “the incruciations having been dismissed”. Tignol [2001, 20] suggests Cardano may have been punning on the word crux (cross), i.e., not only “ignoring the (ex)cruciating pain,” but “dismissing the cross products” involved in the multiplication of the complex binomials.

[17] In 1593, Adriaen van Roomen (1561–1615), a Belgian visitor to the court of Henry IV, challenged the mathematicians of France to solve a 45th degree equation. Viète did so at once, recognizing a term from the expansion of sin(45θ). Viète, a wealthy man, hosted van Roomen at his home for a month, and the two men became friends [Boyer 1985, 341].

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