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: It is enough to prove that $\mathbb Q[\sqrt{2} + \sqrt[3]{3}] =\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ because then the facts below imply that the degree of $\sqrt{2} + \sqrt[3]{3}$ is exactly $6$:


*

*$\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ has degree at least $6$

*$\sqrt{2} + \sqrt[3]{3}$ is a root of a polynomial of degree $6$

*$\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ has degree at most $6$
Alternatively,
$\sqrt{2} + \sqrt[3]{3}$ is a root of $x^6 - 6 x^4 - 6 x^3 + 12 x^2 - 36 x + 1$, which is irreducible because it is irreducible mod $13$. But that's a lot of work to do by hand.
