How can I show that $$[\mathbb Q(\sqrt[3]2,\sqrt[3]5,i\sqrt 3):\mathbb Q(i\sqrt 3)]=9?$$
My idea is: $\sqrt[3]2$ has for its minimal polynpmial $X^3-2$ over $\mathbb Q(i\sqrt 3)$, which I justify by: its roots are given by the elements of the set $$\{\sqrt[3]2,\sqrt[3]2 e^{\frac{2i\pi}{3}},\sqrt[3]2 e^{\frac{4i\pi}{3}}\}$$ and there are no such elements in $\mathbb Q(i\sqrt 3)$. But how can I show that $\sqrt[3]5\notin\mathbb Q(i\sqrt 3, \sqrt[3]2)$ ? Is it enough to show that $\sqrt[3]5\notin\mathbb Q(\sqrt[3]2)$ ? If yes, why, and how would I do it? I thought about supposing that $\sqrt[3]5\in\mathbb Q(\sqrt[3]2)$, so $$5=(a+b\sqrt[3]2+c\sqrt[3]5)^3$$ for certain $a,b,c\in \mathbb Q$ (or $a,b,c\in \mathbb Q(i\sqrt 3)$) and arriving somehow at a contradiction, but the calculation is very long and doesn't give me the contradiction I wanted. Is there another way?