Extension field of $x^3 - 2$ in $\mathbb Q$ I know this has been asked previous on Stackexchange, but I just want to have something clarified:
I am supposed to find an extension K of Q having all roots of $x^3 - 2$ such that $[K:\mathbb Q] = 6.$
Setting $\alpha = 2^\frac{1}{3}$ and $\omega = e^\frac{2\pi i}{3}$, I can write all the roots in a basis such that the degree is $3$: $$\{\alpha,\alpha \omega, \alpha \omega^2\}$$ I guess this is a field of degree 3? So I am then supposed to add three more random linear independent elements such that $[K:\mathbb Q] = 6$?

Remember that $[K:\mathbb{Q}] = [K:\mathbb{Q(\alpha)}][\mathbb{Q}(\alpha):\mathbb{Q}]$. Now you took a third root of unit $\xi$, and $p(x) = x^2 + x +  1 $ is irreducible in $\mathbb{Q}$ which has $\xi$ as root. 
From this what is the degree $[\mathbb Q (\xi): \mathbb Q]$? What does this tell you about $[K:\mathbb{Q(\alpha)}]$?
What you conclude from that?
