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: 14

Answer

$x\sqrt[3] {y^2}$.

Work Step by Step

The given expression is $=\sqrt[6] {x^6y^4}$ Rewrite as an exponential expression. $=(x^6y^4)^{\frac{1}{6}}$ Raise each factor in parentheses to the $\frac{1}{3}$ power. $=(x^6)^{\frac{1}{6}}(y^4)^{\frac{1}{6}}$ Use power rule. $=x^{\frac{6}{6}}y^{\frac{4}{6}}$ Simplify. $=x^1y^{\frac{2}{3}}$ Rewrite in radical notation. $=x\sqrt[3] {y^2}$.
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