# 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: $x^3+b x^2+c x+d=0,\text{ }b,c,d\in\mathbb{C}$ 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 $z^6+q z^3-\frac{p^3}{27}=0.$ This equation is called the cubic resolvent of the reduced cubic $$y^3+p y+q=0,$$ and this can be written as $\left(z^3\right)^2+q z^3-\frac{p^3}{27}=0.$ By the quadratic formula we obtain $z^3=\frac{1}{2}\left(-q\pm \sqrt{\text{\unicode{0008}q}^2+\frac{4p^3}{27}}\right)$ so that $\left.z=\sqrt[3]{\frac{1}{2}\left(-q\pm \sqrt{\text{\unicode{0008}q}^2+\frac{4p^3}{27}}\right.}\right) \\$ 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 $$z=0$$?