Showing that $\sqrt[3]{2}\notin\Bbb Q(\alpha_1,...,\alpha_k)$ where $\alpha_i^2\in\Bbb Q\ \forall i$
Solution 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 $\sqrt[3]{2}\not\in K$.