Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$? I know that $\frac{1}{\sqrt{3+\sqrt{3}}} = \frac{\sqrt{3-\sqrt{3}}}{\sqrt6}$. So I just need to know whether $\sqrt6 \in L$. Since $\sqrt 3 = (\sqrt{3+\sqrt{3}})^2 - 3$, I only need to know if $\sqrt 2 \in L$. I would guess it is not, but how to show it? If I suppose that $\sqrt 2 \in L$ and aim for a contradiction:
It is clear that $\mathbb{Q}(\sqrt 3) \subset L$, and if $\sqrt 2 \in L$, then $\mathbb Q(\sqrt 2) \subset L$ as well. Then $K = \mathbb Q (\sqrt 2, \sqrt 3) \subset L$. Since both $K$ and $L$ are degree $4$ over $\mathbb Q$, this would imply they are equal. But then I'm back at square one.