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: 10

Answer

$2x\sqrt[3]{3}-\sqrt[3]{x^2}$

Work Step by Step

RECALL: (1) Distributive Property: For any real numbers a, b, and c, $a(b-c)=ab-ac$ (2) For any real numbers real numbers a and b within the domain, $\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{ab}$ Use rule (1) above to obtain: $=\sqrt[3]{x}(\sqrt[3]{24x^2}) - \sqrt[3]{x} \cdot \sqrt[3]{x}$ Use rule (2) above to obtain: $=\sqrt[3]{24x^3} - \sqrt[3]{x^2}$ Factor the first radicand so that at least one factor is a perfect cube to obtain: $=\sqrt[3]{8x^3(3)} - \sqrt[3]{x^2} \\=\sqrt[3]{(2x)^3(3)} - \sqrt[3]{x^2}$ Simplify to obtain: $=2x\sqrt[3]{3}-\sqrt[3]{x^2}$
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.