## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\dfrac{\sqrt[4]{28a^2}}{4ab}$
Rationalizing the denominator of the given expression, $\sqrt[4]{\dfrac{7}{64a^2b^4}} ,$ we find: \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{7}{64a^2b^4}\cdot\dfrac{4a^2}{4a^2}} \\\\= \sqrt[4]{\dfrac{28a^2}{256a^4b^4}} \\\\= \dfrac{\sqrt[4]{28a^2}}{\sqrt[4]{256a^4b^4}} \\\\= \dfrac{\sqrt[4]{28a^2}}{\sqrt[4]{(4ab)^4}} \\\\= \dfrac{\sqrt[4]{28a^2}}{4ab} .\end{array} * Note that it is assumed that all variables represent positive numbers.