Answer
$3\sqrt{6}-3$
Work Step by Step
We lose the square roots in the denominator by applying the difference of squares formula:
$(a\sqrt{x}+b\sqrt{y})(a\sqrt{x}-b\sqrt{y})=(a\sqrt{x})^{2}-(b\sqrt{y})^{2}=a^{2}x-b^{2}y$
$\displaystyle \frac{15}{\sqrt{6}+1} \displaystyle \color{red}{ \cdot\frac{\sqrt{6}-1}{\sqrt{6}-1} }\qquad$ (rationalize)
$ =\displaystyle \frac{15(\sqrt{6}-1)}{(\sqrt{6})^{2}-1} =\frac{15(\sqrt{6}-1)}{6-1} =\frac{15(\sqrt{6}-1)}{5}$
$=3(\sqrt{6}-1)$
$= 3\sqrt{6}-3$
