Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $v$?

share|improve this question
    
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
4  
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
1  
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
2  
@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

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
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, $t=u+v=\sqrt[3]{u^3}+\sqrt[3]{v^3}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.