Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 61

$\dfrac{18+6\sqrt{2}}{7}$

Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $ \dfrac{6}{3-\sqrt{2}} ,$ is \begin{array}{l}\require{cancel} \dfrac{6}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}} \\\\= \dfrac{6(3)+6(\sqrt{2})}{3^2-(\sqrt{2})^2} \\\\= \dfrac{18+6\sqrt{2}}{9-2} \\\\= \dfrac{18+6\sqrt{2}}{7} .\end{array}