## Precalculus (6th Edition)

$\color{blue}{\dfrac{\sqrt[3]{4}}{2}}$
RECALL: (1) For nonnegative real numbers $a$ and $b$, $\sqrt[n]{a}\cdot \sqrt{b} = \sqrt[n]{ab}$. (2) For nonnegative real numbers $a$ and $b$ where $b\ne0$, $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$. (3) For nonnegative odd integers, $\sqrt[n]{a^n} = a$ (4) $a^m\cdot a^n=a^{m+n}$ Use rules (1) and (4) above to obtain: $=\dfrac{\sqrt[3]{16m^{2+2}n^3}}{\sqrt[3]{32m^4n^3}} \\=\dfrac{\sqrt[3]{16m^4n^3}}{\sqrt[3]{32m^4n^3}}$ Use rule (2) above to obtain: $\require{cancel}=\sqrt[3]{\dfrac{16m^4n^3}{32m^4n^3}} \\=\sqrt[3]{\dfrac{\cancel{16}\cancel{m^4n^3}}{\cancel{32}^2\cancel{m^4n^3}}} \\=\sqrt[3]{\dfrac{1}{2}}$ Rationalize the denominator by multiplying $\sqrt[3]{2^2}$ to both the numerator and the denominator to obtain: $=\sqrt[3]{\dfrac{1(2^2)}{2(2^2)}} \\=\sqrt[3]{\dfrac{4}{2^3}} \\=\color{blue}{\dfrac{\sqrt[3]{4}}{2}}$