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 - Page 445: 67

Answer

$\dfrac{7}{2\sqrt[3]{49}}$

Work Step by Step

Rationalizing the numerator of $ \sqrt[3]{\dfrac{7}{8}} $ results to \begin{array}{l} \sqrt[3]{\dfrac{7}{8}} \cdot \sqrt[3]{\dfrac{7^2}{7^2}} \\\\= \sqrt[3]{\dfrac{7^3}{8\cdot7^2}} \\\\= \dfrac{7}{2\sqrt[3]{49}} .\end{array}
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