TITLE: Six numbers such that their cubics are just 2 numbers, then these are roots of quadratic polynomial
QUESTION [0 upvotes]: I guess this question is either stupid or well-known (or both), but as you can see from my titel of the question I cannot really formulate it in a way, Google would understand.
I read the introduction to Algebra by Bosch. And very briefly:
Say you have a cubic polynomial (in this case $X^3+aX-b\in\mathbb{Q}[X]$, but I dont know if thats relevant) with roots $x_1,x_2,x_3$. Then we can look at the six numbers 
$x_{\pi(1)}+\zeta x_{\pi(2)}+\zeta^2 x_{\pi(3)}$ for all permutations $\pi\in S_3$, where $\zeta$ is a primitive third root of unity.
One easily verifies, that $\left(x_{\pi(1)}+\zeta x_{\pi(2)}+\zeta^2 x_{\pi(3)}\right)^3$ has only two different values. In fact for all $\pi\in A_3$ the value of that cube is the same, say $\alpha$ and for all $\pi\in S_3\setminus A_3$  we get the same value, say $\beta$.
Thus "clearly" $\alpha$ and $\beta$ are roots of a quadratic polynomial with coefficients in $\mathbb{Q}$.
And sorry, but I don't see the last step. Why should that be true? It feels like thats easy to proof with Galois theory, but I am looking for a elementary reason (since there should be one).

REPLY [1 votes]: To explain the intuition which makes this "obvious" ...
Note that $\alpha+\beta$ and $\alpha \beta$ are invariant under the action of any permutation of the roots - because such permutations either exchange $\alpha$ and $\beta$ or fix them both.
The symmetric expressions in the roots of the polynomial are precisely the expressions fixed under the whole action of the symmetric group (every permutation). If an expression is so fixed it is symmetric, and if it is symmetric it is fixed. Such expressions are rational combinations of the coefficients of the polynomial (which are the elementary symmetric expressions in the roots). If the coefficients of the polynomial are rational so are the elements fixed under the action of the symmetric group.
Such is the situation here.
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