Solving equations by radicals Let $\zeta$ be a complex number, $\zeta\neq 1, \zeta^3=1$. Then the expression
$$(x_1+\zeta x_2+\zeta^2x_3)^3$$
takes only two distinct values when we permute $x_j$'s. $\bf{Why\  this?}$
Hence it (the expression above) satisfies a quadratic equation. $\bf{Why\  again?}$
Hence the sums
$$x_{\pi(1)}+\zeta x_{\pi(2)}+\zeta^2x_{\pi(3)}$$
where $\pi$ runs over all permutations of $\{1,2,3\}$, can be obtained solving a quadratic equation, and then taking a cubic root. Moreover, $x_1,x_2,x_3$ can be obtained by mean of elementary operations from those sums and then the cubic equation
$$x^3+ax=b$$
can be solved in terms of radicals.
Is there anybody so kind to (try to) explain me what's going on here? Thanks in advance
 A: A priori, the expression $(x_1+x_2\zeta+x_3\zeta^2)$ could take on up to $6$ different values when we permute the $x_i$ (because $S_3$ has order $6$).  Note, however, that cyclic permutations of the $x_i$ simply multiply the whole expression by $\zeta$ or $\zeta^2$.  Since $\zeta^3=1$, the expression $(x_1+x_2\zeta+x_3\zeta)^3$ is therefore invariant under the   $C_3$ part of the $S_3$-action.  This leaves at most two values for the expression $(x_1+x_2\zeta+x_3\zeta)^3$, e.g $\alpha_1$ and $\alpha_2$ (dependent on $x_1,x_2,x_3$).
Secondly, you ask why $\alpha_1,\alpha_2$ satisfy a degree two equation.  This is because the polynomial
$$F(X)=(X-\alpha_1)(X-\alpha_2)$$
is invariant under the interchange of $\alpha_1$ and $\alpha_2$.  Consequently, it is invariant under the whole of the $S_3$-action discussed before.  By Galois theory, it must lie in $\mathbb{Q}[X]$, because it certainly had coefficients in the field $\mathbb{Q}(\zeta)$, and is invariant under $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$.
A: Let $f_{\zeta}(x_1,x_2,x_3)= (x_1 + \zeta x_2 + \zeta^2 x_3)^3$
\begin{align}
f_{\zeta}(x_1,x_2,x_3) & = (x_1 + \zeta x_2 + \zeta^2 x_3)^3  = (\zeta^3x_1 + \zeta x_2 + \zeta^2 x_3)^3\\
& = (\zeta(\zeta^2x_1 + x_2 + \zeta x_3))^3 = (\zeta^2x_1 + x_2 + \zeta x_3)^3\\
& = f_{\zeta}(x_2,x_3,x_1)
\end{align}
Hence, we have
$$f_{\zeta}(x_1,x_2,x_3) = f_{\zeta}(x_3,x_1,x_2) = f_{\zeta}(x_2,x_3,x_1)$$
$$f_{\zeta}(x_1,x_3,x_2) = f_{\zeta}(x_2,x_1,x_3) = f_{\zeta}(x_3,x_2,x_1)$$
