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# 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.

## Mathematics Subject Classification

12D10*no label found*

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