Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 25

Answer

$\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}=\dfrac{\sqrt[3]{6xy^{2}}}{3x}$

Work Step by Step

$\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}$ Multiply the fraction by $\dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}}$ and simplify: $\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}=\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}\cdot\dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}}=\dfrac{\sqrt[3]{6xy^{2}}}{\sqrt[3]{27x^{3}}}=\dfrac{\sqrt[3]{6xy^{2}}}{3x}$
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