Cubic formula confusion/clarification

Question

So the cubic formula yields ultimately a degree 6 equation which is easily solved as it just involves solving a quadratic in $z^3$ but this gives 6 solutions. My question is at what step in the process can we say that 3 of the solutions are equal to each other (as in 6 solutions boils down to 3)?

Answer 1

For a reduced cubic $$y^3=3py+2q$$

the resolvents are symmetric in roles:

$$u^3+v^3=2q$$

$$uv=p$$

they are conjugate to each other,

$$u=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}} \implies v=\frac{p}{\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}} =\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}$$

One of the roots $y=u+v$ is just a matter of writing order.

Also one list of roots namely

$$y=u\omega_{3}^{k}+v\omega_{3}^{2k}$$

is equivalent to

$$y=u\omega_{3}^{2k}+v\omega_{3}^{k}$$

where $\omega_n=e^{2\pi i/n}$.

There're cyclic properties for polynomial equations. The Galois theory is a group theory about polynomial zeros. For your interests, see also other posts of mine here, here, here and here.