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# variant of Cardano’s derivation

By a linear change of variable, a cubic polynomial over ${\mathbb{C}}$ can be given the form $x^{3}+3bx+c$. To find the zeros of this cubic in the form of surds in $b$ and $c$, make the substitution $x=y^{{1/3}}+z^{{1/3}},$ thus replacing one unknown with two, and then write down identities which are suggested by the resulting equation in two unknowns. Specifically, we get

$\displaystyle y+3(y^{{1/3}}+z^{{1/3}})y^{{1/3}}z^{{1/3}}+z+3b(y^{{1/3}}+z^{{1/% 3}})+c=0.$ | (1) |

This will be true if

$\displaystyle y+z+c=0$ | (2) | ||

$\displaystyle 3y^{{1/3}}z^{{1/3}}+3b=0,$ | (3) |

which in turn requires

$\displaystyle yz=-b^{3}.$ | (4) |

The pair of equations (2) and (4) is a quadratic system in $y$ and $z$, readily solved. But notice that (3) puts a restriction on a certain choice of cube roots.

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Proof

## Mathematics Subject Classification

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