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: 63

Answer

$x\sqrt[5]{8x^3y}$

Work Step by Step

Write 4 as $2^2$ to obtain: $=\dfrac{2x^2y}{\sqrt[5]{2^2x^2y^4}}$ RECALL: For any real number $a$, $\sqrt[5]{a^5}= a$ Rationalize the denominator by multiplying $\sqrt[5]{2^3x^3y}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{2x^2y \cdot \sqrt[5]{2^3x^3y}}{\sqrt[5]{2^2x^2y^4} \cdot \sqrt[5]{2^3x^3y}} \\=\dfrac{2x^2y\sqrt[5]{8x^3y}}{\sqrt[5]{2^5x^5y^5}} \\=\dfrac{2x^2y\sqrt[5]{8x^3y}}{2xy} \\=\dfrac{\cancel{2}x^{\cancel{2}}\cancel{y}\sqrt[5]{8x^3y}}{\cancel{2xy}} \\=x\sqrt[5]{8x^3y}$
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