Answer
$3\sqrt{2}+2\sqrt{3}$
Work Step by Step
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplifed form of given expression, $
\dfrac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}\cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}
\\\\=
\dfrac{\sqrt{6(3)}+\sqrt{6(2)}}{(\sqrt{3})^2-(\sqrt{2})^2}
\\\\=
\dfrac{\sqrt{18}+\sqrt{12}}{3-2}
\\\\=
\dfrac{\sqrt{9\cdot2}+\sqrt{4\cdot3}}{3-2}
\\\\=
\dfrac{\sqrt{(3)^2\cdot2}+\sqrt{(2)^2\cdot3}}{1}
\\\\=
3\sqrt{2}+2\sqrt{3}
.\end{array}
