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: 47

Answer

$3\sqrt[3]{2}$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.