$F=\mathbb{Q}(\alpha_1,\alpha_2, ... , \alpha_n)$, where $\alpha_i^2 \in \mathbb{Q}$ implies $\sqrt[3]{2} \notin F$

Question

Let $F=\mathbb{Q}(\alpha_1,\alpha_2, ... , \alpha_n)$, where $\alpha_i^2 \in \mathbb{Q}$ for $i=1,2,...,n.$ I want to show that $\sqrt[3]{2} \notin F$. I am trying to prove by contradiction assuming $\sqrt[3]{2} \in F$. However, I am not able to complete the proof. How should I obtain a contradiction from the assumption?

Answer 1

Hint:The degree of the field is $2^m$ and the degree of a field wich contains $2^{1\over 3}$ is divisble by $3$.