# Question Regarding Cardano's Formula

In Cardano's derivation of a root of the cubic polynomial $f(X)=X^3+bX+c$ he splits the variable $X$ into two variables $u$ and $v$ together with the relationship that $u+v=X$. From this he finds that $x=u+v$, where

$$u=\sqrt[3]{\frac{-c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$ and $$v=\sqrt[3]{\frac{-c}{2}-\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$

is a root of $f(X)$.

Is there an intuitive explanation for why Cardano splits the variable $X$ into two parts, $u$ and $v$?

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I think it's a very common idea in math to add in some sort of structured free variable so that you can mess around with things, or to replace a 'simple problem' with a more complicated one so that more things can be tried. If there's anything else to it, I don't know. –  mixedmath Jun 13 '12 at 11:28
My guess is that it comes from looking at solutions to quadratic equations, which are always of the form $u+\sqrt{a}$, which can obviously be split up as a sum of two variables $u$ and $v$, where $v=\sqrt{a}$. If you make the substitution $x=u+v$ into a quadratic equation, you can derive the quadratic formula, so I presume Cardano (or Tartaglia, or Scipione or whoever actually solved it) spotted this trick and decided to apply it to cubics. –  Donkey_2009 Jun 13 '12 at 11:57
Incidentally, we can notice that the solutions to the quadratic are then always of the form $u+v$ and $u-v$, so we can try factorizing the quadratic as $x^2+bx+c=(x-u-v)(x-u+v)$ and get to a solution that way. The equivalent trick for cubics is to factorize them as $x^3+bx+c=(x-t-u-v)(x-t-\omega u-\omega^2v)(x-t-\omega^2 u-\omega v)$, where $\omega$ is a complex cube root of unity. Then it turns out that $t=0$, and you get the same expressions for $u$ and $v$ as Cardano did. This is known as the method of Lagrange multipliers. –  Donkey_2009 Jun 13 '12 at 12:03
How I discovered Cardano's proof: a few years ago, I tried to discover how to wolve the cubic myself, and managed to get to the equations $u^3+v^3=-c$ and $3uv=-b$ using a method similar to the one I just described. The correct way to proceed is now to eliminate $v$ and to form and solve a quadratic in $u^3$. Rather than do that, I noticed that these equations meant that $-c-b(u+v)=(u+v)^3$, and then noticed that I was just looking at the original cubic, with an $x=u+v$ substitution. This meant that I could work backwards from that substitution, and reach the solution of the cubic! –  Donkey_2009 Jun 13 '12 at 12:08
@Donkey_2009 You get $t=0$ by design, not by accident, because you shift so that the coefficient of $x^2$ is zero. If you didn't do this, you'd have a non-zero $t$ and could eliminate it by setting $y=x-t$. –  Mark Bennet Jun 13 '12 at 13:00

A somewhat hindsighted explanation is offered by elementary Galois theory. When we are looking for a formula for the roots of $$x^3+ax^2+bx+c=0,$$ we have the generic Galois group $S_3$. [The generic case treats $a,b,c$ as algebraically independent transcendentals over, say $K=\mathbb{Q}(\sqrt{-3})$ - a base field with enough roots of unity, and studies the splitting field of the cubic polynomial over $K(a,b,c)$.] Cardano knew the elementary trick (roughly equivalent to solving the quadratic by completing the square) of substituting $x=t-(a/3)$. This eliminates the quadratic term, and puts the generic cubic equation in the form $$t^3+pt+q=(t-t_1)(t-t_2)(t-t_3),$$ where $t_1,t_2,t_3$ are the unknown roots. Let $\omega=(-1+i\sqrt3)/2$ be a primitive cubic root of unity. Consider the quantities \begin{aligned} z_1&=t_1+t_2+t_3,\\ z_2&=t_1+\omega t_2+\omega^2 t_3,\\ z_3&=t_1+\omega^2 t_2+\omega t_3. \end{aligned} If we permute the roots according to the 3-cycle: $t_1\mapsto t_2\mapsto t_3\mapsto t_1$, we see that the quantities $z_i, i=1,2,3,$ are multiplied by $1,\omega^2$ and $\omega$ respectively. Therefore their cubes $z_i^3,i=1,2,3,$ are invariant under the action of this 3-cycle. So those cubes belong to a smaller field that must be a quadratic extension over the field of definition [and a general fact tells us that the said quadratic must be $K(a,b,c)(\sqrt{D})$, where $D$ is the discriminant]. Cardano didn't think about it in terms of field extensions, but this explains, why something like his formula must exist.

We observe that it is easy to invert the transformation $F:(t_1,t_2,t_3)\mapsto (z_1,z_2,z_3)$. Many of you hopefully recognize transformation $F$ as the discrete Fourier transform of length 3, but even if you don't, you can solve the linear system and invert the transformation as follows: \begin{aligned} t_1&=\frac13(z_1+z_2+z_3),\\ t_2&=\frac13(z_1+\omega^2 z_2+\omega z_3),\\ t_3&=\frac13(z_1+\omega z_2+\omega^2 x_3). \end{aligned} So knowing $z_i$:s allows us to calculate $t_i$:s (and vice versa). But before we go further, let's take note of the fact that by eliminating that quadratic term, we arrived at a situation, where $z_1=t_1+t_2+t_3=0$, because this is the negative of the coefficient of the quadratic term. Expand the product $(t-t_1)(t-t_2)(t-t_3)$ to see this, if you didn't know this bit in advance.

Getting warmer. With $z_1=0$ no longer a mystery, let's define two more variables and call $u=z_2/3$, $v=z_3/3$. At this point we know that \begin{aligned} t_1&=u+v,\\ t_2&=\omega^2 u+\omega v,\\ t_3&=\omega u+\omega^2 v, \end{aligned} and (if we believe Galois theory) that $u^3$ and $v^3$ belong to a quadratic extension field, and thus must be solvable by the formula for the roots of a certain quadratic extension that we figure out next.

Expanding gives $$t^3+pt+q=(t-(u+v))(t-(\omega^2 u+\omega v))(t-(\omega u+ \omega^2v))= t^3-3uv t-(u^3+v^3).$$ From this we see that $u^3+v^3=-q$ and $u^3v^3=(-p/3)^3$. Therefore $u^3,v^3$ are, indeed, roots of the quadratic equation $$(Y-u^3)(Y-v^3)=Y^2+qY-\frac{p^3}{27}=0.$$ This leads to Cardano's formula.

The trick telling us that the components of the discrete Fourier transformation of the roots, such as the $z_i$:s here, have powers in the fixed field of a cyclic group of permutations of roots, was later systematically exploited in Kummer's theory ($n$ roots and $n^{th}$ roots of unity) characterizing certain type of cyclic field extensions.

Don't know, whether we can call this explanation "intuitive"? Anyway, the $u$ and $v$ can be thought of as the unknown components of the DFT of the vector of roots $(t_1,t_2,t_3)$, and we see a simple case of Kummer theory in action.

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Hoping to answer the question why rather than how Cardano's method works and to lit a path leading to the mystery variables $u$ and $v$. –  Jyrki Lahtonen Jun 14 '12 at 10:46
If I lost you somewhere along the way, then let's summarize: the quantities $u$ and $v$ are such linear combinations of the roots $t_1,t_2,t_3$ of the form $$u=\frac13(t_1+\omega t_2+\omega^2t_3)\quad v=\frac13(t_1+\omega^2 t_2+\omega t_3)$$ that their third powers are stable under the cyclic permutations of the three roots. –  Jyrki Lahtonen Jun 14 '12 at 18:21

Cardano knew that any quadratic equation of the form $$x^2+bx+c=0\tag{1}$$ can be written as $$x^2-(u+v)x+uv=0,\tag{2}$$ where $u$ and $v$ are the roots of the equation. Since by setting $t=u+v$ in the reduced cubic $$t^3+pt+q=0\tag{3}$$ we get $$(u^3+v^3+q)+(3uv+p)(u+v)=0,\tag{4}$$ then every root of the system $$u^3+v^3+q=0\tag{5a}$$ $$3uv+p=0\tag{5b}$$ is a root of $(4)$ as well, and based on the property of the quadratic equation indicated in $(2)$ it's now easy to find a formula for $t$ satisfying equation $(3)$.

Added. We just need to find two numbers $u^3$ and $v^3$ such that their sum is $-q$ and their product is $-p^3/27$, which we know from $(1)-(2)$ are the roots of the quadratic equation $$Y^2+qY-\frac{p^3}{27}=0.\tag{6}$$

Consequently, $t=u+v=\sqrt[3]{u^3}+\sqrt[3]{v^3}$.

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