Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 470: 152

Answer

$6\sqrt{5}-11x\sqrt[3]{5}$

Work Step by Step

Using properties of radicals, the expression $ 3\sqrt{20}-7x\sqrt[3]{40}+3\sqrt[3]{5x^3} $ simplifies to \begin{array}{l} 3\sqrt{4\cdot5}-7x\sqrt[3]{8\cdot5}+3\sqrt[3]{x^3\cdot5} \\= 3\cdot2\sqrt{5}-7x\cdot2\sqrt[3]{5}+3\cdot x\sqrt[3]{5} \\= 6\sqrt{5}-14x\sqrt[3]{5}+3x\sqrt[3]{5} \\= 6\sqrt{5}-11x\sqrt[3]{5} .\end{array}
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