### Instruction: What is the value of $\sqrt{15 - 6\sqrt{6}} + \sqrt{15 + 6\sqrt{6}}$?### Response: This looks like a problem where I might want to rationalize the denominator of some fraction, or maybe use the difference of squares identity.
But first, I want to see if I can simplify the expression inside each square root.
I notice that $15 - 6\sqrt{6} = (3 - \sqrt{6})^2$, since $(3 - \sqrt{6})^2 = 9 - 6\sqrt{6} + 6 = 15 - 6\sqrt{6}$.
Similarly, $15 + 6\sqrt{6} = (3 + \sqrt{6})^2$, since $(3 + \sqrt{6})^2 = 9 + 6\sqrt{6} + 6 = 15 + 6\sqrt{6}$.
So, I can rewrite the expression as $\sqrt{(3 - \sqrt{6})^2} + \sqrt{(3 + \sqrt{6})^2}$.
Now, I can use the property that $\sqrt{a^2} = |a|$ for any real number $a$.
This means that $\sqrt{(3 - \sqrt{6})^2} = |3 - \sqrt{6}|$ and $\sqrt{(3 + \sqrt{6})^2} = |3 + \sqrt{6}|$.
But, I know that $3 - \sqrt{6}$ and $3 + \sqrt{6}$ are both positive numbers, since $3 > \sqrt{6}$.
So, I can drop the absolute value signs and just write $\sqrt{(3 - \sqrt{6})^2} + \sqrt{(3 + \sqrt{6})^2} = 3 - \sqrt{6} + 3 + \sqrt{6}$.
Finally, I can simplify this by combining like terms and get $3 - \sqrt{6} + 3 + \sqrt{6} = 6$.