# Cardano’s derivation of the cubic formula

To solve the cubic polynomial equation $x^{3}+ax^{2}+bx+c=0$ for $x$, the first step is to apply the Tchirnhaus transformation $x=y-\frac{a}{3}$. This reduces the equation to $y^{3}+py+q=0$, where

 $\displaystyle p$ $\displaystyle=$ $\displaystyle b-\frac{a^{2}}{3}$ $\displaystyle q$ $\displaystyle=$ $\displaystyle c-\frac{ab}{3}+\frac{2a^{3}}{27}$

The next step is to substitute $y=u-v$, to obtain

 $(u-v)^{3}+p(u-v)+q=0$ (1)

or, with the terms collected,

 $(q-(v^{3}-u^{3}))+(u-v)(p-3uv)=0$ (2)

From equation (2), we see that if $u$ and $v$ are chosen so that $q=v^{3}-u^{3}$ and $p=3uv$, then $y=u-v$ will satisfy equation (1), and the cubic equation  will be solved!

There remains the matter of solving $q=v^{3}-u^{3}$ and $p=3uv$ for $u$ and $v$. From the second equation, we get $v=p/(3u)$, and substituting this $v$ into the first equation yields

 $q=\frac{p^{3}}{(3u)^{3}}-u^{3}$

which is a quadratic equation in $u^{3}$. Solving for $u^{3}$ using the quadratic formula, we get

 $\displaystyle u^{3}$ $\displaystyle=$ $\displaystyle\frac{-27q+\sqrt{108p^{3}+729q^{2}}}{54}=\frac{-9q+\sqrt{12p^{3}+% 81q^{2}}}{18}$ $\displaystyle v^{3}$ $\displaystyle=$ $\displaystyle\frac{27q+\sqrt{108p^{3}+729q^{2}}}{54}=\frac{9q+\sqrt{12p^{3}+81% q^{2}}}{18}$

Using these values for $u$ and $v$, you can back–substitute $y=u-v$, $p=b-a^{2}/3$, $q=c-ab/3+2a^{3}/27$, and $x=y-a/3$ to get the expression for the first root $r_{1}$ in the cubic formula. The second and third roots $r_{2}$ and $r_{3}$ are obtained by performing synthetic division  using $r_{1}$, and using the quadratic formula on the remaining quadratic factor.

Title Cardano’s derivation of the cubic formula CardanosDerivationOfTheCubicFormula 2013-03-22 12:10:28 2013-03-22 12:10:28 djao (24) djao (24) 12 djao (24) Proof msc 12D10 FerrariCardanoDerivationOfTheQuarticFormula