Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Review: 28

Answer

$\frac{1}{8x^{9}}$

Work Step by Step

$(-2x^{3})^{-3}=(-2)^{-3}\times (x^{3})^{-3}=\frac{1}{(-2)^{3}}\times x^{3\times-3}=\frac{1}{-2\times-2\times-2}\times x^{-9}=\frac{1}{8x^{9}}$
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