## Precalculus (6th Edition)

$\color{blue}{\dfrac{m\sqrt[3]{n^2}}{n}}$
RECALL: (1) For nonnegative real numbers $a$ and $b$, $\sqrt[n]{a}\cdot \sqrt{b} = \sqrt[n]{ab}$. (2) For nonnegative odd integers, $\sqrt[n]{a^n} = a$ (3) $a^m\cdot a^n=a^{m+n}$ Use rule s(1) and (3) above to obtain: $=\dfrac{\sqrt[3]{m^{1+2}n}}{\sqrt[3]{n^2}} \\=\dfrac{\sqrt[3]{m^3n}}{\sqrt[3]{n^2}}$ Rationalize the denominator by multiplying $\sqrt[3]{n}$ to both the numerator and the denominator to obtain: $=\dfrac{\sqrt[3]{m^3n(n)}}{\sqrt[3]{n^2(n)}} \\=\dfrac{\sqrt[3]{m^3n^2}}{\sqrt[3]{n^3}} \\=\dfrac{\sqrt[3]{m^3n^2}}{n}$ Simplify the numerator by bringing out of the radical sign the cube root of $m^3$ to obtain: $=\color{blue}{\dfrac{m\sqrt[3]{n^2}}{n}}$