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: 61

Answer

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

Work Step by Step

Write 8 as $2^3$ to obtain: $=\dfrac{6}{\sqrt[5]{2^3x^3}}$ RECALL: For any real number $a$, $\sqrt[5]{a^5}= a$ Rationalize the denominator by multiplying $\sqrt[5]{2^2x^2}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{6 \cdot \sqrt[5]{2^2x^2}}{\sqrt[5]{2^3x^3} \cdot \sqrt[5]{2^2x^2}} \\=\dfrac{6\sqrt[5]{4x^2}}{\sqrt[5]{2^5x^5}} \\=\dfrac{6\sqrt[5]{4x^2}}{2x} \\=\dfrac{3\cancel{6}\sqrt[5]{4x^2}}{\cancel{2}x} \\=\dfrac{3\sqrt[5]{4x^2}}{x}$
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