Algebra: A Combined Approach (4th Edition)

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

Chapter 10 - Review - Page 746: 28



Work Step by Step

Step 1: We need to transform 27 into a number of power 3 so that exponential powers can cancel each other out Step 2: We find that $27=3^{3}$ Step 3: Therefore, we can write $(-\frac{1}{27})^{\frac{1}{3}}$ as $(-\frac{1}{3^{3}})^{\frac{1}{3}}$ Step 4: $(-\frac{1}{3^{3}})^{\frac{1}{3}}$= $-\frac{1^{\frac{1}{3}}}{3^{3\times\frac{1}{3}}}$ Step 5: $-1^{\frac{1}{3}}$=$-1$ and ${3^{3\times\frac{1}{3}}}$=$3$ as both 3s cancel each other out Step 6: Therefore, what remains is $-\frac{1}{3}$
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