Question:
What is the value of $\sqrt{15 - 6\sqrt{6}} + \sqrt{15 + 6\sqrt{6}}$?

Answer:
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 $-6$.