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: 55

Answer

$\dfrac{7\sqrt[3]{4x}}{2x}$

Work Step by Step

RECALL: For any real number $a$, $\sqrt[3]{a^3}= a$ Rationalize the denominator by multiplying $\sqrt[3]{2^2x}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{7 \cdot \sqrt[3]{2^2x}}{\sqrt[3]{2x^2} \cdot \sqrt[3]{2^2x}} \\=\dfrac{7\sqrt[3]{4x}}{\sqrt[3]{2^3x^3}} \\=\dfrac{7\sqrt[3]{4x}}{2x}$
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