Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Review Exercises: 13

Answer

$\frac{1}{z^{15}}$

Work Step by Step

$(z^{-3})^{3}z^{-6}=z^{-3\times3}z^{-6}=z^{-9}z^{-6}=z^{-9+(-6)}=z^{-9-6}=z^{-15}=\frac{1}{z^{15}}$
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