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 - Page 476: 71

Answer

$\dfrac{\sqrt[3]{12}}{3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $ \sqrt[3]{\dfrac{4}{9}} ,$ 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{4}{9}\cdot\dfrac{3}{3}} \\\\= \sqrt[3]{\dfrac{12}{27}} \\\\= \sqrt[3]{\dfrac{12}{(3)^3}} .\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]{12}}{\sqrt[3]{(3)^3}} \\\\= \dfrac{\sqrt[3]{12}}{3} .\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.