In the splitting field of $x^3 - 2$ show $ \alpha w \in \mathbb{Q}(\alpha^2 + \alpha^2 w)$ 
How do I show $ \alpha w \in \mathbb{Q}(\alpha^2 + \alpha^2 w)$ and similar questions such as 
$ \alpha + \alpha w \in \mathbb{Q}(\alpha^2 + \alpha^2 w)$
 A: This is the same situation as Example 2 on here https://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory#Example_2
The subgroup structure of the Galois group tells us about the lattice of subfields of our field.
Let $t = \alpha^2 + \alpha^2 \omega$ and consider the action of the Galois group on this element:

*

*$\sigma t = \alpha^2 \omega^2 + \alpha^2 \omega^3 = t$

*$\tau t = \alpha^2 + \alpha^2 \omega^2$
Thus $\mathbb Q(t)$ is the fixed field of $\mathbb Q(\alpha,\omega)$ under the automorphism group $\langle \sigma \rangle$ which has order $3$, index $2$.
The Galois Correspondence theorem tells us this will correspond to a degree $3$ field extension, there are $3$ of these so let us try to find which one we have.
We can calculate $$t^2 = (\alpha^2 + \alpha^2 \omega)^2 = \alpha^4 + 2 \alpha^4 \omega + \alpha^4 \omega^2 = 2 \alpha (1 + 2 \omega + \omega) = 2 \alpha \omega$$
So we could have immediately answered the question without any theory just by squaring $t^2/2 = \alpha \omega$.
But now we can also demonstrate that $\alpha + \alpha w \not \in \mathbb Q(t)$ since if it were $\alpha$ would be and thus $\omega$ would be which would imply $\mathbb Q(t) = \mathbb Q(\alpha, \omega)$ but this is not the case.
