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

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

or, with the terms collected,

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

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

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.