# variant of Cardano’s derivation

By a linear change of variable, a cubic polynomial over $\u2102$ 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

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

This will be true if

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

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

which in turn requires

$yz=-{b}^{3}.$ | (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.

Title | variant of Cardano’s derivation |
---|---|

Canonical name | VariantOfCardanosDerivation |

Date of creation | 2013-03-22 13:38:43 |

Last modified on | 2013-03-22 13:38:43 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 9 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 12D10 |