In an attempt to solve $x^3+px=-q$
Note,
$$(a-b)^3+3ab(a-b)=a^3-b^3$$
Let $p=3ab$ and $-q=a^3-b^3$. Then we have,
$$(a-b)^3+p(a-b)=-q$$
$a-b$ is a solution to the cubic equation $x^3+px=-q$. If we assume $p \neq 0$ then we have $b=\frac{p}{3a}$, otherwise solving the depressed cubic is simple. It is important to note that if $p \neq 0$ then neither $a$ nor $b$ can be zero.
We substitute $b=\frac{p}{3a}$ in to $-q=a^3-b^3$ and eventually get,
$$a^3=\frac{-q \pm \sqrt{q^2+\frac{4}{27}p^3}}{2}$$
That gives $6$ possible choices for $a$ (at most) and $6$ possible choices for $b$. I would like to finish, and find $a-b$.
My Question:
May someone please explain why this way of Cardano's method gives only $3$ different solutions instead of what appears to be $6$.