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I have a general question about the cubic equation:

Let $a,b,c \in \mathbb{C}$ with $a \neq 0$. The general cubic equation is $t^3+at^{2}+bt+c = 0$. To get Cardano's Formula, we first transform the equation so that $a= 0$ (i.e. $t^3+bt+c = 0$). Then we reduce this to a quadratic that we can solve. What is the motivation behind the transformations that reduce $t^3+at^2+bt+c = 0$ to $t^3+bt+c = 0$?

In particular, if we let $y = t+ \frac{a}{3}$, then $t = y-\frac{a}{3}$ which cancels the $at^2$ term. This is called a Tschirnhaus transformation. We are left with an equation of the form $$y^3+py+q = 0$$

But how do we know what transformations to use to get Cardano's Formula, where $$p = \frac{a^2-2a^3+3b}{3}\quad\mathrm{and}\quad q = \frac{2a^3-9ab+27c}{27}\quad ?$$ Finally if we let $y = \sqrt[3]{u}+ \sqrt[3]{v}$ we eventually get a quadratic which leads to Cardano's Formula.

J. M. ain't a mathematician
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Damien
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    If you want motivation, I think Lagrange's derivation of the cubic formula is much better motivated (the motivation comes from Galois theory): http://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method – Qiaochu Yuan Jul 06 '11 at 21:10
  • Yes, sorry about that. – Qiaochu Yuan Jul 06 '11 at 22:03
  • I haven't read Tschirnhaus, have Cardano. Dissecting cubes was used by him to get what we would now think of as identities for homogeneous cubics. May have been motivated by similar strategies for quadratics that go back to al-Khwarizmi. The elimination of $x^2$ was presumably motivated by the success of del Ferro in solving at least one non-trivial class of cubics with the $x^2$ term missing. – André Nicolas Jul 06 '11 at 23:47
  • @user: +1. To be more specific, del Ferro solved the classes of equations $x^3=px+q$ and $x^3+q=px$ for positive (integers) $p$ and $q$. – Did Jul 07 '11 at 06:13
  • @Didier Piau: Thanks, I thought the speculation was $x^3+px=q$ with uncertainty about $x^3=px+q$. Worried about this a number of years ago, couldn't locate anything but very second-hand stuff. – André Nicolas Jul 07 '11 at 06:56
  • @Didier Piau: Didn't del Ferro solve $x^3 = px+q$, $x^3 + px= q$, and $x^3+q = px$? – Damien Jul 07 '11 at 20:33
  • @Damien What do you mean by the "motivation" for the change of variable $t = y - a/3$? Do you mean (1) why do we want to eliminate the quadratic term, or (2) how do we know that that substitution will do that? – Jack M Jun 21 '14 at 16:18
  • Just FYI, the expression for $p$ shown here is incorrect; see https://math.stackexchange.com/q/4443786/326036 – Anakhand May 05 '22 at 15:11

3 Answers3

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I don't think it's really all that mysterious. The motivation behind using the Tschirnhaus Transformation seems clear; it's better to have fewer terms and fewer coefficients when solving an equation. It next probably occurred to Cardano to try $y = u + v$ and see what happens. Rewrite the equation as $$u^3 + v^3 + 3uv(u + v) + p(u + v) + q = 0$$ You have a little leeway in choosing $u$ and $v$, since you only need them to sum to a root of the equation. So you might want to impose a second condition on $u$ and $v$ to simplify things (and two equations in the two variables $u$ and $v$ should specify their values). To reduce the above to a quadratic equation, you can stipulate that $uv = -{p \over 3}$. Namely you have $$u^3 + v^3 + q = 0$$ which is equivalent to $$u^3 -{p^3 \over 27}{1 \over u^3} + q = 0$$ Or just $$u^6 + qu^3 - {p^3 \over 27} = 0$$ This is a quadratic equation in $u^3$, which you can solve. $u$ can be any of the three cube roots of this, and $v$ is determined by the equation $3uv = -p$. Thus you have three possible values of $u + v$ which will be the three roots of your equation.

In Cardano's time they didn't really know much about complex numbers (i.e. that every number has $n$ complex $n$th roots) but he was smart enough to basically understand what was going on.

Zarrax
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  • Cardano barely mentioned complex numbers, knew they were absurd and that he had no clue, and stopped. About a quarter century later, Bombelli, who also did not know what was going on, did some formal manipulations of what we now call complex numbers, in the same cubics context. – André Nicolas Jul 07 '11 at 02:12
  • He was able to take the real cube roots and obtain a real root in that case. I think also he had some idea that since a cubic always has a real root $u$ and $v$ could always be complex conjugates if they weren't real. But not anything that precise or rigorous. – Zarrax Jul 07 '11 at 02:37
  • On reading Ars Magna, you will find Cardano did not know about complex conjugates. There is a only a fleeting reference to what we would call complex numbers, followed by immediate dismissal. (Later, in his work on the casus irreducibilis, Bombelli did manipulate complex conjugates.) – André Nicolas Jul 07 '11 at 03:07
  • How do we justify that we can always write $y$ as $u + v$ with the additional condition that $u^3 + v^3 + q = 0$? Otherwise the method would only apply to such cases where that is indeed possible, right? – Anakhand May 05 '22 at 15:29
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From a historical viewpoint, Lagrange's method of resolvents was certainly not the first method to solve cubics, but it is much more intelligible from a "modern" viewpoint, as here . Oddly, many "modern" treatments of Galois theory do not show how to solve any equations by radicals, and, in particular, omit Lagrange's method. Years ago, van der Waarden's "Algebra" was one of the few popular texts to do so.

In the specific case of cubics, Lagrange's method can be viewed as approximately recapitulating Cardano's manipulations, and similarly for quartics.

Lagrange's method also succeeds in cyclotomic extensions and some other explicit abelian extensions.

paul garrett
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That bugs me a little too, but I think the great advantage is this one: by making $t=u+v$, $y=t+k$ or whatever, you are giving your problem more degrees of freedom. Since there is more than one tuple $(u,v)$ that satisfies this relation, you will be able to impose another condition on the new coefficients in order to make the problem similar to something you can solve. By annulating one of them, for example.

It is also interesting to think the other way around. At first you don't know how to solve $ax^2+bx+c=0$ (I). However, you can always solve the simpler equation $x^2-p=0$ (II). If you can prove that by substituting $x$ for $[mx+n]$ in (II) you can always arrive at (I) for appropriate $m$, $n$ and $p$, you'll also be able to solve (I). Problem is: in this case we have 3 equations - one for comparing each coefficient - and three variables. As the degree gets bigger*, we keep having three variables but the number of equations increases, yielding the possibility of no solutions to the system.

*$a_0+a_1 x+\dots+a_k x^k=0 = (mx+n)^k-p=0$

Dario
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user159378
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