Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.4 - Adding, Subtracting, and Dividing Radical Expressions - Exercise Set - Page 538: 32

Answer

$\dfrac{\sqrt[3]{11}}{4}$

Work Step by Step

RECALL: The quotient rule: $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ where $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$ Use the quotient rule above to obtain: $=\dfrac{\sqrt[3]{11}}{\sqrt[3]{64}} \\=\dfrac{\sqrt[3]{11}}{\sqrt[3]{4^3}} \\=\dfrac{\sqrt[3]{11}}{4}$
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