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

$\dfrac{\sqrt[5]{48a^4b^3}}{2b^{2}}$
Rationalizing the denominator of the given expression, $\dfrac{\sqrt[5]{3a^4}}{\sqrt[5]{2b^7}} ,$ we find: \begin{array}{l}\require{cancel} \dfrac{\sqrt[5]{3a^4}}{\sqrt[5]{2b^7}}\cdot\dfrac{\sqrt[5]{16b^3}}{\sqrt[5]{16b^3}} \\\\= \dfrac{\sqrt[5]{48a^4b^3}}{\sqrt[5]{32b^{10}}} \\\\= \dfrac{\sqrt[5]{48a^4b^3}}{\sqrt[5]{(2b^{2})^5}} \\\\= \dfrac{\sqrt[5]{48a^4b^3}}{2b^{2}} .\end{array} * Note that it is assumed that all variables represent positive numbers.