Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 27

Answer

$\dfrac{\sqrt[3]{6xy^2}}{3x}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{\sqrt[3]{2y^2}}{\sqrt[3]{9x^2}} $ results to \begin{array}{l} \dfrac{\sqrt[3]{2y^2}}{\sqrt[3]{9x^2}} \cdot \dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}} \\\\= \dfrac{\sqrt[3]{6xy^2}}{\sqrt[3]{27x^3}} \\\\= \dfrac{\sqrt[3]{6xy^2}}{3x} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.