Is it true $\mathbb Q(\sqrt{2},\sqrt[3]{2})=\mathbb Q(\sqrt{2}+\sqrt[3]{2})$?

Question

Is it true $\mathbb Q(\sqrt{2},\sqrt[3]{2})=\mathbb Q(\sqrt{2}+\sqrt[3]{2})$? If yes prove it if not explain why

I can show that $\mathbb Q(\sqrt{2}+\sqrt[3]{2})$ ⊂ $\mathbb Q(\sqrt{2},\sqrt[3]{2})$ but im not sure that $\mathbb Q(\sqrt{2},\sqrt[3]{2})$ ⊂ $\mathbb Q(\sqrt{2}+\sqrt[3]{2})$.

Answer 1

They both have degree 6 over $\mathbb{Q}$ and one field is contained in the other. By the tower law they are the same.