Finding a primitive element for a field extension

Question

Find a primitive element for the field extension $\mathbb Q(2^{1/3},2^{1/4})$ of $\mathbb Q$

Answer 1

Approach 1: Notice that $\mathbb Q(2^{1/3},2^{1/4})=\mathbb Q(2^{1/12})$ this is immediate from writing $2^{1/3}2^{1/4}=2^{7/12}=\sqrt{2}2^{1/12}$ so $2^{1/12} \in \mathbb Q(2^{1/3},2^{1/4})$. It follows that $2^{1/12}$ is a primitive element.

Approach 2: Notice that the Galois closure of $\mathbb Q(2^{1/3},2^{1/4})$ is $\mathbb Q(2^{1/3},2^{1/4},i,\omega)$ where $\omega$ is a primitive third root of unity. Show that $2^{1/3}+2^{1/4}$ is a primitive element because it's not fixed by any automorphism that doesn't fix $2^{1/3}$ or $2^{1/4}$.