#### Answer

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

#### Work Step by Step

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}