Answer
$\dfrac{6-x^2}{8\sqrt[]{6}-8x}$
Work Step by Step
Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt[]{6}+x}{8}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt[]{6}+x}{8}\cdot\dfrac{\sqrt[]{6}-x}{\sqrt[]{6}-x}
\\\\=
\dfrac{(\sqrt[]{6})^2-(x)^2}{8\sqrt[]{6}-8x}
\\\\=
\dfrac{6-x^2}{8\sqrt[]{6}-8x}
.\end{array}
