I've been studying Cardano's method lately, and there's one step that I haven't been able to convince myself is correct.
The method relies two times on the following fact:
Lemma: For $a, b\in \Bbb C$, there exists a unique pair $u, v$ with $u + v=a$ and $uv=b$.
We start with a depressed monic cubic having some root $x$:
$$x^3+ax+b=0$$
Obviously there exists $u, v$ with $u+v=x$, and when we expand with the binomial formula and gather terms, we can obtain:
$$u^3+v^3+b+(u+v)(3uv+a)=0$$
By our lemma, there exists a unique pair such that on top of $u+v=x$, we also have $uv=-a/3$. In other words, the following system (1) holds:
$$\begin{align}u+v&=x\\ uv&=-a/3\end{align}\tag{1}$$
For that pair, we get (2):
$$\begin{align}u^3+v^3&=-b\\ u^3v^3&=-a^3/27\end{align}\tag{2}$$
By our lemma again, we can get expressions $u^3$ and $v^3$. It is then claimed that $u$ and $v$ are gotten by taking the cube roots of these expressions. That's the step that seems unclear to me, especially considering each one has $3$ cube roots anyway.
Why is it that if $u, v$ satisfy system (2), they must satisfy system (1)? Or, do the cube roots in the cubic formula refer to a specific choice of cube root? If so, how do you choose the correct one?
You have three choices for $u$ and independently three choices for $v$ to satisfy $u^3+v^3=-b$ and $u^3v^3=-a^3/27$. But for each choice of $u$, you have no more choice for $v$ to satisfy $uv=-a/3$. So ultimately you have only three choices,whic is fine as there are three roots.