## Intermediate Algebra (6th Edition)

$2x^{2}\sqrt[4]{5y^{3}}$
Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the given expression, $\dfrac{\sqrt[4]{160x^{10}y^5}}{\sqrt[4]{2x^2y^2}} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{160x^{10}y^5}{2x^2y^2}} \\= \sqrt[4]{80x^{10-2}y^{5-2}} \\= \sqrt[4]{80x^{8}y^{3}} .\end{array} Extracting the factor of the radicand that is a perfect power of the index, the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{16x^{8}\cdot5y^{3}} \\= \sqrt[4]{(2x^{2})^4\cdot5y^{3}} .\end{array} Using $\sqrt[n]{x^n}=|x|$ if $n$ is even and $\sqrt[n]{x^n}=x$ if $n,$ the expression above is equivalent to \begin{array}{l}\require{cancel} |2x^{2}|\sqrt[4]{5y^{3}} \\= 2|x^{2}|\sqrt[4]{5y^{3}} \\= 2x^{2}\sqrt[4]{5y^{3}} .\end{array}