jmc

Solution 1: Let $x=\sqrt{15-6\sqrt{6}}+\sqrt{15+6\sqrt{6}}.$ Then $$x^2={\left(\sqrt{15-6\sqrt{6}}\right)}^2+2\sqrt{15-6\sqrt{6}}\sqrt{15+6\sqrt{6}}+{\left(\sqrt{15+6\sqrt{6}}\right)}^2$$ We observe that $\left(15-6\sqrt{6}\right)\left(15+6\sqrt{6}\right)=15^2-{\left(6\sqrt{6}\right)}^2=225-216=9$ because of difference of squares. So $$x^2=\left(15-6\sqrt{6}\right)+2\sqrt{9}+\left(15+6\sqrt{6}\right)$$ The $6\sqrt{6}$ terms cancel, and so $x^2=36.$ Since $x$ must be positive, then $x=\boxed{6}$ and not $-6$.

Solution 2: Let $a+b\sqrt{6}=\sqrt{15+6\sqrt{6}}$ for some $a$ and $b$. Squaring, we get $(a^2+6b^2)+2ab\sqrt{6}=15+6\sqrt{6}$. After some experimentation, we see this is true if $a=3$, $b=1$. So $\sqrt{15+6\sqrt{6}}=3+\sqrt{6}$. Similarly, we find that $\sqrt{15-6\sqrt{6}}=3-\sqrt{6}$. So $\sqrt{15-6\sqrt{6}}+\sqrt{15+6\sqrt{6}}=(3-\sqrt{6})+(3+\sqrt{6})=\boxed{6}$.