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 - Page 549: 7


$\sqrt[3]{12} + 4\sqrt[3]{10}$

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: $=\sqrt[3]{2}(\sqrt[3]{6}) + \sqrt[3]{2} \cdot 4\sqrt[3]{5}$ Use rule (2) above to obtain: $=\sqrt[3]{2(6)} + 4\sqrt[3]{2(5)} \\=\sqrt[3]{12} + 4\sqrt[3]{10}$
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