I am trying to prove that $K=\mathbb{Q}(2^{1/3}, i\sin{2\pi/3})$ is Galois extension over $\mathbb{Q}$. It is easy to see that $K=\mathbb{Q}(2^{1/3},i\sqrt{3})$. I know it is Galois since $K$ is a splitting field of the separable polynomial $f(x)=x^3-2$.
Now I am trying to show this using the other method, that is by explicitly computing the automorphisms. I found that $|Aut(E/F)|=6$. However I am having a hard time proving $[K:\mathbb{Q}]=6$. In particular I am having a hard time finding an irreducible degree 2 polynomial which will say that $[K:\mathbb{Q}(2^{1/3})]=2$.
Once you have proven that $Aut(K/\mathbb{Q})$ is of cardinal $6$, you just need to know that $[K:\mathbb{Q}]$ is below $6$, which is an easy thing to do because :
$$[K:\mathbb{Q}]=[K:\mathbb{Q}(2^{\frac{1}{3}})][\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}] $$
We easily get that :
$$[\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}]=3$$
And :
$$[K:\mathbb{Q}(2^{\frac{1}{3}})]=1\text{ or } 2$$
because $x^2+3$ is a polynomial with coefficients in $\mathbb{Q}(2^{\frac{1}{3}})$ and $i\sqrt{3}$ is one of its root (for the inequality you do not need to show it is irreducible). Finally :
$$[K:\mathbb{Q}]\leq 6 $$
Now : $6=|Aut(K/\mathbb{Q})|\leq [K:\mathbb{Q}]\leq 6$ which allows you to say that $[K:\mathbb{Q}]= 6$.