Polynomial roots. Finite Extensions. 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 $λ'=1+2ω\sqrt[3]{2}-\bar ω\sqrt[3]{4}$, $λ''=1+2 \bar ω\sqrt[3]{2}-ω\sqrt[3]{4}$.
 A: You largely cannot avoid explicit (tedious) calculation.  You are at various times determining whether two vector spaces intersect, determining whether an element from a vector space is in a subvector space, et c.
In this particular case, you ask whether a particular algebraic number will always appear as a root with some other algebraic number.  Note that conjugate roots (relative to a base field) always occur together as roots of any polynonmial (with coefficients in that base field).  For instance, of polynomials with coefficients in $\Bbb{Q}$, $\sqrt{2}$ and $-\sqrt{2}$ are conjugate roots and always appear together.  You might already know that a minimal polynomial contains the conjugate roots.  (The Galois group of the splitting field of the minimal polynomial transitively maps the algebraic number you care about to all the other roots of the minimal polynomial.)
So, can you find the minimal polynomial of $1+2 \sqrt[3]{2} - \sqrt[3]{4}$?
(It's $x^3 - 3x^2+15x-25$.)
Can you divide this root from that and factor the result?
(The quotient is $x^2 - (2-2\sqrt[3]{2} + \sqrt[3]{4})x + 5+5\sqrt[3]{2}$ and factoring quadratics is straightforward.  This is where $\omega$ comes into the picture because these roots are not real and $1 \pm \mathrm{i}\sqrt{3}$ will be handy in representing them.)
I'm not justifying any of that because it should be straightforward...  If not, let me know.
A: 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.
