Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Cumulative Review Exercises - Page 872: 20


The answer is $\frac{5\sqrt[3] {4xy^2}}{2xy}$.

Work Step by Step

The given expression is $\frac{5}{\sqrt[3] {2x^2y}}$ To rationalize the denominator we multiply and divide by $\sqrt[3] {2^2xy^2}$ $\frac{5}{\sqrt[3] {2x^2y}} \cdot \frac{\sqrt[3] {2^2xy^2}}{\sqrt[3] {2^2xy^2}}$ Simplify. $\frac{5\sqrt[3] {2^2xy^2}}{2xy}$. $\frac{5\sqrt[3] {4xy^2}}{2xy}$.
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