Intermediate Algebra (6th Edition)

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

Chapter 10 - Cumulative Review - Page 634: 27c

Answer

$14xy^2\sqrt[3]{x}$

Work Step by Step

Using the properties of radicals, the expression $ \dfrac{7\sqrt[3]{48x^4y^8}}{\sqrt[3]{6y^2}} $ simplifies to \begin{array}{l} 7\sqrt[3]{\dfrac{48x^4y^8}{6y^2}} \\\\= 7\sqrt[3]{8x^4y^{8-2}} \\\\= 7\sqrt[3]{8x^4y^{6}} \\\\= 7\sqrt[3]{8x^3y^{6}\cdot x} \\\\= 7\cdot2xy^2\sqrt[3]{x} \\\\= 14xy^2\sqrt[3]{x} .\end{array}
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