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: 50

Answer

$\dfrac{\sqrt[3]{6}}{2}$

Work Step by Step

RECALL: For any real number $a$, $\sqrt[3]{a^3}= a$ Rationalize the denominator by multiplying $2$ to both the numerator and the denominator of the radicand. Simplify using the rule above to obtain: $\require{cancel} =\sqrt[3]{\dfrac{3(2)}{4(2)}} \\=\sqrt[3]{\dfrac{6}{8}} \\=\sqrt[3]{\dfrac{6}{2^3}} \\=\frac{1}{2}\sqrt[3]{6} \\=\dfrac{\sqrt[3]{6}}{2}$
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