Answer
$\dfrac{3\sqrt[4]{2x^2}}{2x^{3}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the denominator of the given expression, $
\sqrt[4]{\dfrac{81}{8x^{10}}}
,$ simplify first the radicand by extracting the root of the factor that is a perfect power of the index. Then multiply by an expression equal to $1$ which will make the denominator a perfect power of the index. Then extract again the root of the factor that is a perfect power of the index.
$\bf{\text{Solution Details:}}$
Extracting the root of the factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
\sqrt[4]{\dfrac{81}{x^{8}}\cdot\dfrac{1}{8x^2}}
\\\\=
\sqrt[4]{\left( \dfrac{3}{x^{2}}\right)^4\cdot\dfrac{1}{8x^2}}
\\\\=
\dfrac{3}{x^{2}}\sqrt[4]{\dfrac{1}{8x^2}}
.\end{array}
Multiplying the given expression by an expression equal to $1$ which will make the denominator a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{3}{x^{2}}\sqrt[4]{\dfrac{1}{8x^2}\cdot\dfrac{2x^2}{2x^2}}
\\\\=
\dfrac{3}{x^{2}}\sqrt[4]{\dfrac{1}{16x^4}\cdot2x^2}
.\end{array}
Extracting the root of the factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{3}{x^{2}}\sqrt[4]{\left(\dfrac{1}{2x}\right)^4\cdot2x^2}
\\\\=
\dfrac{3}{x^{2}}\cdot\dfrac{1}{2x}\sqrt[4]{2x^2}
\\\\=
\dfrac{3}{2x^{3}}\sqrt[4]{2x^2}
\\\\=
\dfrac{3\sqrt[4]{2x^2}}{2x^{3}}
.\end{array}
