See my previous question for a refresher on how Cardano's method for the cubic works. The gist is that where $x$ is any root of the cubic, we know that $x = u+v$, where we have explicit formulae for $u^3$ and $v^3$. So now we want to take cube roots of $u^3$ and $v^3$, and add them together to form $x$.
We can't just take any cube roots. However, what we can do is take an arbitrary cube root of $u^3$, and then since we know:
$$uv=-n/3$$
we can conclude that $v=\frac{-n/3}{u}$. But how do we know $u\neq0$?
I get that we can suppose, since the expressions involving $u$ and $v$ are symmetric, that $u$ is non zero as long as we can guarantee that at least one of $u$ or $v$ is non-zero. I can also see that if both $u$ and $v$ are zero, then the cubic is just $x^3=0$, which is trivial. But still, the cubic formula is supposed to be general. Not only that, but just plugging in zero coefficients to the cubic formula does yield a correct solution, so clearly this division by zero isn't actually a problem. Why is that?
Since, by construction, $u^3v^3 = -a^3/27$, the only way $u$ (or $v$) can be zero is if $a=0$. However, if $a=0$ then the initial cubic equation was just $x^3+b=0$, and then solving it is just a matter of taking a single cube root -- no need for Cardano in the first place.