Algebra: A Combined Approach (4th Edition)

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

Chapter 5 - Section 5.2 - Negative Exponents and Scientific Notation - Exercise Set: 47

Answer

$\frac{27a^{6}}{b^{12}}$

Work Step by Step

We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer). Therefore, $(3a^{2}b^{-4})^{3}=3^{3}\times(a^{2})^{3}\times(b^{-4})^{3}=27\times a^{2\times3}\times b^{-4\times3}=27\times a^{6}\times b^{-12}=27\times a^{6}\times \frac{1}{b^{12}}=\frac{27a^{6}}{b^{12}}$.
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