Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set: 41

Answer

$\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}=\dfrac{\sqrt[3]{2}}{6}$

Work Step by Step

$\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}$ Rewrite the first term as $\dfrac{\sqrt[3]{8\cdot2}}{\sqrt[3]{27}}$ and the second term as $\dfrac{\sqrt[3]{27\cdot2}}{6}$: $\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}=\dfrac{\sqrt[3]{8\cdot2}}{\sqrt[3]{27}}-\dfrac{\sqrt[3]{27\cdot2}}{6}=...$ Simplify both terms: $...=\dfrac{2\sqrt[3]{2}}{3}-\dfrac{3\sqrt[3]{2}}{6}=\dfrac{2\sqrt[3]{2}}{3}-\dfrac{\sqrt[3]{2}}{2}=...$ Evaluate the substraction and simplify if possible: $...\Big(\dfrac{2}{3}-\dfrac{1}{2}\Big)\sqrt[3]{2}=\Big(\dfrac{4-3}{6}\Big)\sqrt[3]{2}=\dfrac{\sqrt[3]{2}}{6}$
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