variant of Cardano’s derivation

By a linear change of variable, a cubic polynomial over $ℂ$ canbe given the form $x^3+3bx+c$. To find the zeros of this cubic in theform 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)

$3y^{1/3}{z}^{1/3}+3b=0,$(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 certainchoice of cube roots.