Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$? 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.

Let $\alpha, \beta$ be $\sqrt{3+\sqrt{3}}$ and $\sqrt{3-\sqrt{3}}$ respectively; they clearly have the same minimum polynomial on $\mathbb{Q}$ which is $f(x) = x^4-6x^2+6$, by direct calculation and using Eisenstein Criterion.
Now let $E$ be the splitting field of $f$ on $\mathbb{Q}$ and let $G$ be its Galois group. The order of $G$ may be $4,8,12,24$; I'll prove that it is exactly 8.
In this way it can be shown that $\beta$ cannot be in $\mathbb{Q}(\alpha)$ because otherwise $E = \mathbb{Q}(\alpha)$ which has order 4 on $\mathbb{Q}$.
By Galois Correspondence Theorem there is at least one subgroup $H$ of $G$ with index 4 in it: that's the subgroup consisting of all $\mathbb{Q}$-automorphisms of E which fix the elements of $\mathbb{Q}(\alpha)$.
Since the elements of H send roots of $f$ in roots of $f$, it's clear that the only possibility for $H$ is to be of the form $H = \{id, \sigma\}$ where $\sigma$ is the $\mathbb{Q}$-automorphism of E that sends $\beta$ to $-\beta$ and fixes $\alpha$ (and consequently fixes $-\alpha$).
Therefore $G$ contains a subgroup having 2 elements and index 4 in $G$; by Lagrange's Theorem follows that $G$ has order 8.
