degree of $\sqrt{2} + \sqrt[3]{3}$ over $\mathbb Q$ How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?
Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2\sqrt{3} = (\sqrt{3} - \sqrt{2}) + (\sqrt{2} + \sqrt{3})$ etc.
But how can i express $\sqrt{2}$ or $\sqrt[3]{3}$ with $\sqrt{2} + \sqrt[3]{3}$?
Is $\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ equal to $\mathbb Q[\sqrt{2}, \sqrt[3]{3}]$?
Thanks

Hint:
Check which automorphisms of ${\mathbf Q}[\sqrt 2,\sqrt[3]{3},e^{2\pi i/3}]$ fix ${\mathbf Q}[\sqrt2+\sqrt[3]{3}]$.
