Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review - Page 469: 90

Answer

$3a\sqrt[4]{2a}$

Work Step by Step

Using the properties of radicals, the expression $ 3\sqrt[4]{32a^5}-a\sqrt[4]{162a} $ simplifies to \begin{array}{l} 3\sqrt[4]{16a^4\cdot2a}-a\sqrt[4]{81\cdot2a} \\= 3\cdot2a\sqrt[4]{2a}-a\cdot3\sqrt[4]{2a} \\= 6a\sqrt[4]{2a}-3a\sqrt[4]{2a} \\= 3a\sqrt[4]{2a} .\end{array}
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