I want to show $\mathbb{Q}(\sqrt{3+\sqrt{3}})/\mathbb{Q}$ is not a normal extension and conclude that the normal closure is $\mathbb{Q}(\sqrt{3+\sqrt{3}},\sqrt{3-\sqrt{3}})$. After knowing the former, the latter is easy to show.
This entire thing boils down to showing that $\sqrt{3-\sqrt{3}}\notin \mathbb{Q}(\sqrt{3+\sqrt{3}})$, and since $$\left(\sqrt{3+\sqrt{3}}\right)\cdot\left(\sqrt{3-\sqrt{3}}\right)=\sqrt{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}=\sqrt{9-6}=\sqrt{6}$$ and $\sqrt{3}\in \mathbb{Q}(\sqrt{3+\sqrt{3}})$ all that's left to show is that $\sqrt{2}\notin \mathbb{Q}(\sqrt{3+\sqrt{3}})$ but I couldn't find a way to do that.
EDIT: I am looking for an elegant solution that doesn't use norms, resultants or algebra that is too dirty. The posts mentioned in the comments are not satisfying to me