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: 56

Answer

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

Work Step by Step

RECALL: For any real number $a$, $\sqrt[3]{a^3}= a$ Rationalize the denominator by multiplying $\sqrt[3]{2x}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{10 \cdot \sqrt[3]{2x}}{\sqrt[3]{4x^2} \cdot \sqrt[3]{2x}} \\=\dfrac{10\sqrt[3]{2x}}{\sqrt[3]{8x^3}} \\=\dfrac{10\sqrt[3]{2x}}{\sqrt[3]{2^3x^3}} \\=\dfrac{10\sqrt[3]{2x}}{2x} \\=\dfrac{5\cancel{10}\sqrt[3]{2x}}{\cancel{2}x} \\=\dfrac{5\sqrt[3]{2x}}{x}$
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.