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: 46

Answer

$\dfrac{\sqrt[3]{9}}{3}$

Work Step by Step

RECALL: For any real number $a$, $\sqrt[3]{a^3}= a$ Rationalize the denominator by multiplying $\sqrt[3]{3^2}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{1 \cdot \sqrt[3]{3^2}}{\sqrt[3]{3} \cdot \sqrt[3]{3^2}} \\=\dfrac{\sqrt[3]{3^2}}{\sqrt[3]{3^3}} \\=\dfrac{\sqrt[3]{9}}{3}$
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