Answer
$=\frac{8a^7}{9b^{11}}$
Work Step by Step
$\frac{(3a^{-2}b)^{-2}}{(2ab^{-3})^{-3}}$
$=\frac{3^{-2}a^{4}b^{-2}}{2^{-3}a^{-3}b^{9}}$
$=\frac{1}{3^2}(2^3)(a^{4-(-3)})(b^{-2-9})$
$=\frac{2^3}{3^2}(a^7)(b^{-11})$
$=\frac{8a^7}{9b^{11}}$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 9 - Cumulative Review - Page 677: 32
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