Intermediate Algebra (6th Edition)

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

Chapter 7 - Test - Page 470: 13

Answer

$\dfrac{\sqrt[3]{b^2}}{b}$

Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the denominator a perfect power of the radical, the rationalized-denominator form of the given expression, $ \dfrac{\sqrt[3]{ab}}{\sqrt[3]{ab^2}} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt[3]{ab}}{\sqrt[3]{ab^2}}\cdot\dfrac{\sqrt[3]{a^2b}}{\sqrt[3]{a^2b}} \\\\= \dfrac{\sqrt[3]{a^3b^2}}{\sqrt[3]{a^3b^3}} \\\\= \dfrac{\sqrt[3]{(a)^3\cdot b^2}}{\sqrt[3]{(ab)^3}} \\\\= \dfrac{a\sqrt[3]{b^2}}{ab} \\\\= \dfrac{\cancel{a}\sqrt[3]{b^2}}{\cancel{a}\cdot b} \\\\= \dfrac{\sqrt[3]{b^2}}{b} .\end{array}
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