Answer
$4\sqrt{6}+8$
Work Step by Step
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $
\dfrac{8}{\sqrt{6}-2}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{8}{\sqrt{6}-2}\cdot\dfrac{\sqrt{6}+2}{\sqrt{6}+2}
\\\\=
\dfrac{8(\sqrt{6}+2)}{(\sqrt{6})^2-(2)^2}
\\\\=
\dfrac{8(\sqrt{6}+2)}{6-4}
\\\\=
\dfrac{8(\sqrt{6}+2)}{2}
\\\\=
\dfrac{\cancel{2}(4)(\sqrt{6}+2)}{\cancel{2}}
\\\\=
4(\sqrt{6}+2)
\\\\=
4\sqrt{6}+8
.\end{array}
