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: 12

Answer

$\dfrac{5\sqrt[3]{9y^2}}{3y}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{5}{\sqrt[3]{3y}} $ results to \begin{array}{l} \dfrac{5}{\sqrt[3]{3y}} \cdot \dfrac{\sqrt[3]{9y^2}}{\sqrt[3]{9y^2}} \\\\= \dfrac{5\sqrt[3]{9y^2}}{\sqrt[3]{27y^3}} \\\\= \dfrac{5\sqrt[3]{9y^2}}{3y} .\end{array}
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