Algebra: A Combined Approach (4th Edition)

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

Chapter 5 - Section 5.6 - Special Products - Exercise Set: 88

Answer

$\frac{-3b^{3}}{2}$

Work Step by Step

$\frac{-48ab^{6}}{32ab^{3}}=\frac{-48a^{1-1}b^{6-3}}{32}=\frac{-48a^{0}b^{3}}{32}=\frac{-48(1)b^{3}}{32}= \frac{-3b^{3}}{2}$
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