Find :
$a.[\mathbb{Q(\sqrt{3 + \sqrt{2}})} : \mathbb{Q}]$.
$b.[\mathbb{Q(\sqrt{3}+ 2^\frac{1}{3})} : \mathbb{Q}]$.
My attempt:
$a. x= (\sqrt{3 + \sqrt{2}}) \implies x^2=3+ \sqrt{2} \implies x^4-6x^2+7=0 \implies [\mathbb{Q(\sqrt{3 + \sqrt{2}})} : \mathbb{Q}] = 4$
b.$x=\sqrt{3}+ 2^\frac{1}{3}$ , I try to get a minimal polynomial $p$ such that $p(x)=\sum_{i=0}^n a_ix^i=0$ , such that $a_i \in \mathbb{Q}$.
Is $(a)$ is correct (I am not entirely sure that this is the right wat to solve this problem) ?
Any idea how to solve $(b)$ ?