## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 550: 71

#### Answer

$\displaystyle \frac{3\sqrt[4]{x^{3}y}}{x^{2}y}$

#### Work Step by Step

In the denominator, apply $\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab},\qquad$ and $\sqrt[n]{a^{n}}=(\sqrt[n]{a})^{n}=a$ (for positive a). $\sqrt[4]{x^{5}y^{3}}=\sqrt[4]{x^{4}\times xy^{3}}=\sqrt[4]{x^{4}}\times\sqrt[4]{xy^{3}}=x\sqrt[4]{xy^{3}}$ we rationalize so that we get $\sqrt[4]{x^{4}y^{4}}$=$xy$ in the numerator $\displaystyle \frac{3}{x\sqrt[4]{xy^{3}}} \displaystyle \color{red}{ \cdot\frac{\sqrt[4]{x^{3}y}}{\sqrt[4]{x^{3}y}} }\qquad$ (rationalize) $=\displaystyle \frac{3\sqrt[4]{x^{3}y}}{x\sqrt[4]{x^{4}y^{4}}}$ $=\displaystyle \frac{3\sqrt[4]{x^{3}y}}{x\times xy}$ $=\displaystyle \frac{3\sqrt[4]{x^{3}y}}{x^{2}y}$

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