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).
Compare https://en.wikipedia.org/wiki/Cubic_function#Cardano%27s_method
I got $$ \alpha \beta = -27 a^3$$ while $$ \alpha + \beta = 9 b $$ so that $$ (\alpha - \beta)^2 = 81 b^2 + 108 a^3 $$ and we can take $\pm$ square root in the last one.
First try, I need to get groceries, I will probably go through it again later. The basic observation, that you have enough information to write both $\alpha \beta$ and $\alpha + \beta$ in terms of $a,b,$ is correct.
There is an explicit part about symmetric functions involved: your roots sum to zero. Then we have (I will change letters) $$ p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2+q^2+r^2 - qr - rp - pq),$$ $$ p^2+q^2+r^2 - qr - rp - pq = (p+q+r)^2 - 3(qr+rp+pq) $$