$\mathbb{Q}(\sqrt[3]{2},e^{\frac{2\pi i}{3}})=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}e^{\frac{2\pi i}{3}})$

Question

I have to show that those two field extensions are the same, however I'm struggling to get the job done.

$\mathbb{Q}(\sqrt[3]{2},e^{\frac{2\pi i}{3}})=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}e^{\frac{2\pi i}{3}})$

Answer 1

Since $\sqrt[3] 2 , e^{\frac {2 \pi i} {3}} \in \Bbb Q (\sqrt[3] 2 , e^{\frac {2 \pi i} {3}})$, so $\sqrt[3] 2 e^{\frac {2 \pi i} {3}} \in \Bbb Q(\sqrt[3] 2 , e^{\frac {2 \pi i} {3}})$. So $\Bbb Q(\sqrt[3] 2 , \sqrt[3] 2e^{\frac {2 \pi i} {3}}) \subseteq \Bbb Q(\sqrt[3] 2 , e^{\frac {2 \pi i} {3}})$ because $\Bbb Q (\sqrt[3] 2 , \sqrt[3] 2 e^{\frac {2 \pi i} {3}})$ is the smallest field containing $\Bbb Q, \sqrt[3] 2$ and $\sqrt[3] 2 e^{\frac {2 \pi i} {3}}$. Can you show the other inclusion?