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 - Test - Page 578: 8

Answer

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

Work Step by Step

Simplify. $\sqrt[3]{16x^4y^8}$ Therefore, $\sqrt[3]{16x^4y^8}=\sqrt[3]{8x^3y^6} \cdot \sqrt[3] {2xy^2}$ or, $=\sqrt[3]{2^3x^3(y^2)^3} \sqrt[3] {2xy^2}$ Thus, the given radical term can be written as:$2xy^2\sqrt[3] {2xy^2}$
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