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 - Mid-Chapter Check Point - Page 541: 6

Answer

$4xy^{\frac{1}{12}}$.

Work Step by Step

The given expression is $=(64x^3y^{\frac{1}{4}})^{\frac{1}{3}}$ Raise each factor in parentheses to the $\frac{1}{3}$ power. $=(4^3)^{\frac{1}{3}}(x^3)^{\frac{1}{3}}(y^{\frac{1}{4}})^{\frac{1}{3}}$ Use power rule. $=4^{\frac{3}{3}}x^{\frac{3}{3}}y^{\frac{1}{4\cdot3}}$ Simplify. $=4xy^{\frac{1}{12}}$.
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