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 - Review Exercises - Page 577: 56

Answer

$5\sqrt[3]{2}-\sqrt[3]{10}$.

Work Step by Step

The given expression is $=\sqrt[3]5(\sqrt[3]{50}-\sqrt[3]{2})$ Use the distributive property. $=\sqrt[3]5(\sqrt[3]{50})-\sqrt[3]5(\sqrt[3]{2})$ Use product rule. $=\sqrt[3]{5\cdot50}-\sqrt[3]{5\cdot2}$ Simplify. $=\sqrt[3]{250}-\sqrt[3]{10}$ Factor as a perfect cube. $=\sqrt[3]{5^3\cdot 2}-\sqrt[3]{10}$ Use product rule. $=\sqrt[3]{5^3}\cdot \sqrt[3]{2}-\sqrt[3]{10}$ Simplify. $=5\sqrt[3]{2}-\sqrt[3]{10}$.
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