Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 469: 89

Answer

$9\sqrt[3]{2}$

Work Step by Step

Using the properties of radicals, the expression $ \sqrt[3]{128}+\sqrt[3]{250} $ simplifies to \begin{array}{l} \sqrt[3]{64\cdot2}+\sqrt[3]{125\cdot2} \\= 4\sqrt[3]{2}+5\sqrt[3]{2} \\= 9\sqrt[3]{2} .\end{array}
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