Cardano's Formula from a^3+b^3+c^3-3abc

Is it possible to solve the cubic using the factorization of $u^3+v^3+w^3 - 3uvw$?

Say we have a cubic $P(x) = ax^3+bx^2+cx+d$ (with $a\ne0$), and we want to find its roots $r,s,t$. Is the factorization $$ax^3+bx^2+cx+d = a(x-r)(x-s)(x-t)$$ related to the factorization $$u^3+v^3+w^3 - 3uvw = (u+v+w)(u+\zeta v + \zeta^2 w)(u + \zeta^2 v + \zeta^4 w),$$ where $\zeta = e^{2\pi i/3}$ is a primitive third root of unity?

Challenge to the community: Use these ideas to give an alternative proof of Cardano's formula.

We wish to solve $ax^3+bx^2+cx+d=0$ using the factorization given above. Let $r$, $s$, and $t$ be the roots of our cubic, so that $x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=\left(x-r\right)\left(x-s\right)\left(x-t\right)$ Equating coefficients of powers of $x$ gives: $-r-s-t=\frac{b}{a}$ so $r+s+t=-\frac{b}{a}$

$rs+rt+st=\frac{c}{a}$

$-rst=\frac{d}{a}$ so $rst=-\frac{d}{a}$

Now we make a helpful substitution: $r=u+v+w$

$s=u+\zeta v+\zeta^2w$

$t=u+\zeta^2v+\zeta w$,

where, as above, $\zeta=\frac{-1+\sqrt{-3}}{2}$ is a primitive cube root of unity. Note that the product of the coefficients on $v$ and $w$ is always 1; this will be important at the end. (The matrix of this substitution is a Vandermonde matrix, which is invertible because $1$, $\zeta$, and $\zeta^2$ are distinct complex numbers. That is why this substitution can be made.)

The first equation immediately gives us $u$: $r+s+t=3u=-\frac{b}{a}$, so $u=-\frac{b}{3a}$

The second equation then becomes $rs+rt+st=3u^2-3vw=\frac{c}{a}$ so, since $u=-\frac{b}{3a}$, $vw=\frac{b^2-3ac}{9a^2}$

The third equation becomes the identity: $rst=u^3+v^3+w^3-3uvw=-\frac{d}{a}$

Substituting in $u=-\frac{b}{3a}$ gives $-\frac{b^3}{27a^3}+v^3+w^3+\frac{b}{a}vw=-\frac{d}{a}$ so that $v^3+w^3+\frac{b}{a}vw=\frac{b^3-27a^2d}{27a^3}$

Remembering that $vw=\frac{b^2-3ac}{9a^2}$, we substitute that in and we can obtain $v^3+w^3=\frac{9abc-27a^2d-2b^3}{27a^3}$

This means that $v^3$ and $w^3$ are the roots of the quadratic $\left(\tau-v^3\right)\left(\tau-w^3\right)=\tau^2-\left(v^3+w^3\right)\tau+v^3w^3=\tau^2-\frac{\left(9abc-27a^2d-2b^3\right)}{27a^3}\tau+\frac{\left(b^2-3ac\right)^3}{729a^6}$

Then the solutions to our cubic are $\frac{-b+\sqrt[3]{\tau_1}+\sqrt[3]{\tau_2}}{3a}$, where $\tau_1$ and $\tau_2$ are the roots of the above quadratic, and the cube roots are taken in pairs whose product is always $(3a\zeta^{-k}v)(3a\zeta^{k}w)=(3av)(3aw)=9a^2vw= b^2-3ac$