Let $L:=\mathbb{Q}(\omega, \sqrt[3]{2})$ with $\omega:=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+\frac{i}{2}\sqrt{3}$.
How to find a primitive element of the field extension $\mathbb{Q} \subset L$?
My idea was to show that $\mathbb{Q}(\omega, \sqrt[3]{2})=\mathbb{Q}(\omega+\sqrt[3]{2})$
So I started with:
$\omega + \sqrt[3]{2}\in \mathbb{Q}(\omega, \sqrt[3]{2}) \Rightarrow \mathbb{Q}(\omega+\sqrt[3]{2}) \subset \mathbb{Q}(\omega,\sqrt[3]{2})$
For the other direction I got:
$(\omega+\sqrt[3]{2})^3=3+3(-1)^{\frac{2}{3}}2^{\frac{2}{3}}-3(-1)^{\frac{1}{3}}\sqrt[3]{2}$
$\Rightarrow \sqrt[3]{2}\in \mathbb{Q}(\omega+ \sqrt[3]{2})$
Now I don't know how to continue to show that $\omega \in \mathbb{Q}(\omega+ \sqrt[3]{2})$.
What has to be done next?
Since you proved that $\sqrt[3]{2} \in \mathbb{Q}(\omega+\sqrt[3]{2})$, then $$ \omega=(\omega+\sqrt[3]{2})-\sqrt[3]{2} \in \mathbb{Q}(\omega+\sqrt[3]{2}) $$ and you are done.