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 \begin{eqnarray*} p & = & b - \frac{a^2}{3} \\ q & = & c - \frac{ab}{3} + \frac{2a^3}{27} \end{eqnarray*}The next step is to substitute $y = u-v$ , to obtain \begin{equation}\label{original} (u-v)^3 + p(u-v) + q = 0 \end{equation}or, with the terms collected, \begin{equation}\label{rearrange} (q - (v^3 - u^3)) + (u-v)(p - 3 u v) = 0 \end{equation}From equation (), 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 (), 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 \begin{eqnarray*} u^3 & = & \frac{-27q + \sqrt{108 p^3 + 729 q^2}}{54} = \frac{-9q + \sqrt{12 p^3 + 81 q^2}}{18}\\ v^3 & = & \frac{27q + \sqrt{108 p^3 + 729 q^2}}{54} = \frac{9q + \sqrt{12 p^3 + 81 q^2}}{18} \end{eqnarray*}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.