Intermediate Algebra (6th Edition)

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

Chapter 8 - Cumulative Review - Page 533: 6c

Answer

$\frac{16b^{6}}{a^{6}}$

Work Step by Step

$(\frac{3a^{8}b^{2}}{12a^{5}b^{5}})^{-2}$ =$\frac{(12)^{2}(a^{-16})(b^{-4})}{(3)^{2}(a^{-10})(b^{-10})}$ =$\frac{144(b^{-4+10})}{9(a^{-10+16})}$ =$\frac{16(b^{6})}{(a^{6})}$ =$\frac{16b^{6}}{a^{6}}$
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