TITLE: Polynomial roots. Finite Extensions.
QUESTION [1 upvotes]: If $a,b,c$ belong to $\mathbb Q$ and $ω^3=1$ is a third root of unity, then prove that if $$(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3=1+2\sqrt[3]{2}-\sqrt[3]{4}$$ holds we also have 
$$ 1+2ω\sqrt[3]{2}-\bar ω\sqrt[3]{4}=(a+bω\sqrt[3]{2}+c\bar ω\sqrt[3]{4})^3
.$$ (Not sure about the third power on the right hand side  notes are not clear.)  
How can I avoid brute force calculation in things like this? They pop everywhere in field extensions - Galois theory. If a polynomial over $\mathbb Q$ has a root $λ=1+2\sqrt[3]{2}-\sqrt[3]{4}$ then prove it has also the roots $λ'=1+2ω\sqrt[3]{2}-\bar ω\sqrt[3]{4}$, $λ''=1+2 \bar ω\sqrt[3]{2}-ω\sqrt[3]{4}$.

REPLY [1 votes]: Let $K=\mathbb Q(\sqrt[3] 2,\zeta_3)$ which is the splitting field of $x^2 - 2$ over $\mathbb Q$. Since that polynomial is irreducible, it follows that there is an automorphism of $K$ over $\mathbb Q$ which takes $\sqrt[3] 2$ to $\zeta_3 \sqrt[3] 2$. Call it $\sigma$. For clarity let $\sqrt[3] 2 = \alpha$ and $\zeta = \zeta_3$, so $\sigma(\alpha) = \zeta\alpha$.
We know that
$$(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3=1+2\sqrt[3]{2}-\sqrt[3]{4}$$
Equivalently:
$$ a + b \alpha + c\alpha^2 = 1 + 2 \alpha - \alpha^2$$
Since $\sigma$ is an automorphism (thus well defined, and also making the relationship if and only if), it should preserve this equality:
$$\Leftrightarrow \sigma(a + b \alpha + c\alpha^2 = 1 + 2 \alpha - \alpha^2) = \sigma(a + b \alpha + c\alpha^2 = 1 + 2 \alpha - \alpha^2)$$
Distribute $\sigma$ according to the ring homomorphism rules, and recognize that it fixes $a,b,c$ because they are in $\mathbb Q$.
$$\Leftrightarrow a + b \sigma(\alpha) + c (\sigma(\alpha))^2 = 1 + 2\sigma(\alpha) - (\sigma(\alpha))^2$$
Now recognize that $\sigma \alpha = \zeta\alpha = \omega \sqrt[3] 2$ in your notation. Its square is $\zeta^2 \alpha^2$: but $\zeta^2$ is another third root of unity, and $\zeta^2\zeta = 1$, so $\zeta^2 = \zeta^{-1} = \bar \zeta$, and so we get $\bar \omega \sqrt[4] 3$ exactly as you have.
I think you can see how this generalizes to other situations.
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