Cardan's Cubic Formula

Back when I taught university math, students would remember the quadratic formula and then naturally ask if there was a cubic formula, and how that works. Well here goes: We start with the general cubic equation $a x^3+b x^2+c x+d=0$ and clean it up a bit: $$x^3+b x^2+c x+d=0,\text$$ where $b/a, c/a$, and $d/a$ have been replaced with $b$, $c$, and $d$, respectively. Keep in mind the fact that $x^3+b x^2+c x+d$ is a monic polynomial and that reducing to the monic case has no effect on the roots. Now, multiplying by $z^3$we conclude that $y^3+p y+q=0$ is equivalent to the equation $$z^6+q z^3-\frac=0.$$ This equation is called the cubic resolvent of the reduced cubic $y^3+p y+q=0,$ and this can be written as $$\left(z^3\right)^2+q z^3-\frac=0.$$ By the quadratic formula we obtain $$z^3=\frac\left(-q\pm \sqrt\right)$$ so that $$\left.z=\sqrt[3]\left(-q\pm \sqrt\right) \\ $$ Substituing this gives a root of the reduced cubic $y^3+p y+q$, and then $x=y-b/3$ is a root of the cubic $x^3+b x^2+c x+d$. Question: is this derivation correct, and if not why? By setting $y^3+p y+q=0$, we essentially assumed that a solution exists. What justifies this assumption? A cubic equation has three roots, yet the cubic resolvent has degree 6. Why? We assumed $z\neq 0$, what happens when $z=0$?