Simplify: $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$

Question

Through following a walkthrough on https://en.wikipedia.org/wiki/Nested_radical, I have got $2\sqrt{6}-2\sqrt{3}$. Is my answer correct?

Answer 1

$$\sqrt{9-6\sqrt2}+\sqrt{9+6\sqrt2}=\sqrt{3(3-2\sqrt2)}+\sqrt{3(3+2\sqrt2)}=$$ $$=\sqrt{3(\sqrt2-1)^2}+\sqrt{3(\sqrt2+1)^2}=\sqrt6-\sqrt3+\sqrt6+\sqrt3=2\sqrt6.$$

Answer 2

$$x:=\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\;\;\;/^2$$ $$x^2 = 18+2\sqrt{81-72} = 18+6 =24$$

So $x=\sqrt{24}$, since $x>0$. So it seems that your answer is not correct.

Answer 3

$$9-6\sqrt{2}=3(3-2\sqrt{2})=3(1-2\sqrt{2}+(\sqrt{2})^2)=3(\sqrt{2}-1)^2$$ $$9+6\sqrt{2}=3(3+2\sqrt{2})=3(1+2\sqrt{2}+(\sqrt{2})^2)=3(\sqrt{2}+1)^2$$

$$\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}=\sqrt{3}(\sqrt{2}-1)+\sqrt{3}(\sqrt{2}+1)=2\sqrt{6}$$