It is easy to convert the general cubic into the form $$y^3+py+q=0$$ via the Tschirnhaus transformation. However, afterwards we can derive with a lot of work Cardano's formula: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$
In my book on Galois theory by Stewart, he recalls every nonzero complex number has $3$ cube roots. If one of those cube roots is $\alpha$, then the other two are $\omega\alpha$ and $\omega^2\alpha$ where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. This is fine. However then, Stewart says directly proceeding this: "The expression for $y$ therefore appears to lead to nine solutions, of the form $$\alpha+\beta,\space\space \alpha+\omega\beta,\space\space \alpha+\omega^2\beta$$ $$\omega\alpha+\beta,\space\space \omega\alpha+\omega\beta,\space\space \omega\alpha+\omega^2\beta$$ $$\omega^2\alpha+\beta,\space\space \omega^2\alpha+\omega\beta,\space\space \omega^2\alpha+\omega^2\beta$$ where $\alpha,\beta$ are specific choices of cube roots."
I am confused how he makes the jump from Cardano's formula and a reasoning about every $z\in\mathbb{C}\setminus\{0\}$ having $n$ distinct $n\text{-th}$ roots to this permutation argument.
Stewart further argues that not all of the above expressions are zeros (by the Fundamental Theorem of Algebra) and because we let $\space3\sqrt[3]{u}\sqrt[3]{v}+p=0$ in our derivation of Cardano's formula, we should choose $\alpha,\beta$ so that $3\alpha\beta+p=0,$ then the solutions are $$\alpha+\beta,\space\space \omega\alpha+\omega^2\beta, \space\space\omega^2\alpha+\omega\beta$$
How does Stewart arrive at this permuted list of $9$ solutions and further, how does he narrow them down to the three above?
I hope since these two questions are so very closely related in this section of the book I'll be able to ask them both in the same question here. Thanks!