Cardano's derivation of the cubic formulaCardanosDerivationOfTheCubicFormula2002-01-06 06:53:282005-07-24 03:52:52Proofcubic formula0\usepackage{amssymb}
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\newcommand{\qth}{{\mbox{$q^{\text{th}}$}}}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 \eqref{rearrange}, 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 \eqref{original}, 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.