Algebra: A Combined Approach (4th Edition)

$\dfrac{3\sqrt[4]{2x^2}}{2x^{3}}$
$\bf{\text{Solution Outline:}}$ 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}