Rationalize the denominator. \( \frac{3 \sqrt{2}+\sqrt{10}}{6 \sqrt{2}-\sqrt{10}} \) \( \frac{3 \sqrt{2}+\sqrt{10}}{6 \sqrt{2}-\sqrt{10}}=\square \) (Simplify your answer. Type an exact answer, using radicals as needed.)

To rationalize the denominator of the expression \( \frac{3 \sqrt{2}+\sqrt{10}}{6 \sqrt{2}-\sqrt{10}} \), we can multiply both the numerator and the denominator by the conjugate of the denominator, which is \( 6\sqrt{2} + \sqrt{10} \). Let's perform the multiplication: \[ \frac{(3\sqrt{2} + \sqrt{10})(6\sqrt{2} + \sqrt{10})}{(6\sqrt{2} - \sqrt{10})(6\sqrt{2} + \sqrt{10})} \] First, we calculate the denominator: \[ (6\sqrt{2} - \sqrt{10})(6\sqrt{2} + \sqrt{10}) = (6\sqrt{2})^2 - (\sqrt{10})^2 = 72 - 10 = 62 \] Now, let's find the numerator: \[ (3\sqrt{2} + \sqrt{10})(6\sqrt{2} + \sqrt{10}) = 3\sqrt{2} \cdot 6\sqrt{2} + 3\sqrt{2} \cdot \sqrt{10} + \sqrt{10} \cdot 6\sqrt{2} + \sqrt{10} \cdot \sqrt{10} \] \[ = 18 \cdot 2 + 3\sqrt{20} + 6\sqrt{20} + 10 \] \[ = 36 + 9\sqrt{20} + 10 \] \[ = 46 + 9\sqrt{20} \] We can simplify \( \sqrt{20} = 2\sqrt{5} \), so: \[ = 46 + 9(2\sqrt{5}) = 46 + 18\sqrt{5} \] Thus, we combine our results: \[ \frac{46 + 18\sqrt{5}}{62} \] Finally, we can simplify this fraction by dividing both the numerator and denominator by 2: \[ = \frac{23 + 9\sqrt{5}}{31} \] The final answer is: \[ \frac{3 \sqrt{2}+\sqrt{10}}{6 \sqrt{2}-\sqrt{10}} = \frac{23 + 9\sqrt{5}}{31} \]