## Precalculus (6th Edition)

$\color{blue}{2q}$
Use the product rule $\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$ (since p and q are positive) to obtain: $=\dfrac{\sqrt[4]{(8p^2q^5)(2p^3q)}}{\sqrt[4]{p^5q^2}} \\=\dfrac{\sqrt[4]{16p^5q^6}}{\sqrt[4]{p^5q^2}}$ Use the quotient rule $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$ (since p and q are positive) to obtain: $\require{cancel} =\sqrt[4]{\dfrac{16p^5q^6}{p^5q^2}} \\=\sqrt[4]{\dfrac{16\cancel{p^5}\cancel{q^6}q^4}{\cancel{p^5}\cancel{q^2}}} \\=\sqrt[4]{16q^4} \\=\sqrt[4]{2^4q^4} \\=\color{blue}{2q}$