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.

Use the tracial method, for example. The trace map on a number field $Q(\alpha)$ assigns to each $\beta \in \mathbb Q(\alpha)$ the quantity $\sum_{\sigma} \sigma(\beta)$ where $\sigma(\beta)$ are all the conjugates of $\beta$.
Suppose that $\sqrt 2 \in L$. Then $\sqrt 3 \in L$ implies that $\mathbb Q(\sqrt{3+\sqrt 3})  =  \mathbb Q(\sqrt 2 ,\sqrt 3)$.
Write $\sqrt{3 + \sqrt 3} = a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$. Taking the trace of both sides gives $2a = 0$(note that $\sqrt{3+\sqrt 3}$ has minimal polynomial $x^4 - 6x^2+6$ hence has trace zero) hence $a = 0$.
Next multiply by $\sqrt 2$ to get $2b + c \sqrt 6 + 2d \sqrt 3 = \sqrt{6+2\sqrt 3}$. Once again the trace of the RHS is seen to be zero(find the minimal polynomial, it is easy!), so $b = 0$ is obtained on taking trace. 
We are left with $\sqrt{3+\sqrt 3} = c \sqrt 3 + d \sqrt 6$. Multiplying by $\sqrt 3$ gives $\sqrt{9+3\sqrt 3} = 3c + 2d \sqrt 2$. Again the trace of the LHS is zero(easy again!) so we get $c = 0$. 
Finally we are left with $\sqrt{3 + \sqrt{3}} = d \sqrt 6$, here squaring gives $\sqrt 3 = 6d^2  - 3$, an obvious problem. Hence, we are done.
Note that $\sqrt{3 - \sqrt 3}$ is a conjugate of $\sqrt{3+\sqrt 3}$, therefore the above shows that this extension is not normal, hence not Galois.
