Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 550: 95


$\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$

Work Step by Step

$ \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, $\quad \sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab}$ $=\displaystyle \frac{2x}{\sqrt[3]{2x^{2}y}}$
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