$\mathbb Q$ Field extension

Question

Consider the Field $F = \mathbb Q(2^{\frac 1 3})$, Is $\sqrt 2 \in F$?

I'm trying to figure out how to determine that and similar questions, can you give me a hint or some guidance on how to do that?

Answer 1

Hint:

If $\sqrt{2} \in \mathbb{Q}(2^{⅓})$ then $[\mathbb{Q}(\sqrt{2},2^{⅓}) : \mathbb{Q}(2^{⅓})] = 1 \implies [\mathbb{Q}(2^{⅓},\sqrt{2}) : \mathbb{Q}] = [\mathbb{Q}(2^{⅓}) : \mathbb{Q}] = 3$

Hint 2: (A different approach)

The basis of the vector space $\mathbb{Q}(2^{⅓})$ over $\mathbb{Q}$ is $$ 1,2^{⅓},2^{⅔} $$