Intermediate Algebra for College Students (7th Edition)

$\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$
$\displaystyle \frac{\sqrt[3]{4x}}{\sqrt[3]{}y}= \frac{\sqrt[3]{2^{2}x}}{\sqrt[3]{}y}$ ... We want $\sqrt[3]{2^{3}x^{3}}=2x$ in the numerator, so we rationalize with $\displaystyle \frac{\sqrt[3]{2^{2}x}}{\sqrt[3]{y}}\color{red}{ \cdot\frac{\sqrt[3]{2x^{2}}}{\sqrt[3]{2x^{2}}} }\qquad$ (rationalize) For the denominator, $\quad \sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab}$ $=\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$