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: 64

Answer

$\dfrac{3y\sqrt[5]{4x^4y^2}}{2}$

Work Step by Step

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