Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 73

Answer

$\dfrac{\sqrt[3]{18}}{4}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \sqrt[3]{\dfrac{9}{32}} ,$ multiply the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{9}{2^5}} \\\\= \sqrt[3]{\dfrac{9}{2^5}\cdot\dfrac{2}{2}} \\\\= \sqrt[3]{\dfrac{18}{2^6}} .\end{array} Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt[3]{18}}{\sqrt[3]{2^6}} \\\\= \dfrac{\sqrt[3]{18}}{\sqrt[3]{(2^2)^3}} \\\\= \dfrac{\sqrt[3]{18}}{2^2} \\\\= \dfrac{\sqrt[3]{18}}{4} .\end{array}
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