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.

Title variant of Cardano’s derivation VariantOfCardanosDerivation 2013-03-22 13:38:43 2013-03-22 13:38:43 mathcam (2727) mathcam (2727) 9 mathcam (2727) Proof msc 12D10