Showing that $\sqrt[3]{2}\notin\Bbb Q(\alpha_1,...,\alpha_k)$ where $\alpha_i^2\in\Bbb Q\ \forall i$

Question

Let $F$ be the field $\Bbb Q(\alpha_1,...,\alpha_k)$ where $\alpha_i^2\in\Bbb Q\ \forall i$

I want to show that $\sqrt[3]{2}\notin F$

I thought if $\sqrt[3]{2}\in F$ then $\sqrt[3]{2}\in\Bbb Q\cup(\alpha_1,...,\alpha_k)$ but this is not a disjoint union so I'm not sure what to do... taking the square of $\beta:=\sqrt[3]{2}$ gives $2^{2/3}=\sqrt[3]{4}\notin\Bbb Q$ but that doesn't really advance things

So I'm looking for a hint

Answer 1

$\mathbb{Q}(\sqrt[3]{2})$ has degree $3$ over $\Bbb{Q}$, but the other field $K$ has degree $2^l$ for some $l$. Hence $\mathbb{Q}(\sqrt[3]{2})\subseteq K$ is impossible by the degree theorem. Hence $\sqrt[3]{2}\not\in K$.