Find the degree of the field extension $\mathbb{Q}[\sqrt[3]{2},\sqrt[3]{3}]$ over $\mathbb{Q}$.
My approach: Call the desired degree $n$. Clearly, $3|n$ and $n\leq 9$. So possible values of $n$ are $3,6,9$. I don't know how to go further. I can't show $\sqrt[3]{2} \notin \mathbb{Q}[\sqrt[3]{3}]$, or vice versa. I am thinking of writing $\mathbb{Q}[\sqrt[3]{3}]$ in terms of basis elements and then using contradiction to show $\sqrt[3]{2} \notin \mathbb{Q}[\sqrt[3]{3}]$, but the approach is too ugly to continue.
I don't know Galois Theory.