## Intermediate Algebra for College Students (7th Edition)

The answer is $=-3(\sqrt 3+\sqrt 5)$.
The given expression is $\frac{6}{\sqrt 3-\sqrt 5}$ Divide and multiply by $\sqrt 3+\sqrt 5$ to rationalize the denominator. $=\frac{6}{\sqrt 3-\sqrt 5}\cdot \frac{\sqrt 3+\sqrt 5}{\sqrt 3+\sqrt 5}$ $=\frac{6(\sqrt 3+\sqrt 5)}{3-5}$ $=\frac{6(\sqrt 3+\sqrt 5)}{-2}$ $=-3(\sqrt 3+\sqrt 5)$.