Intermediate Algebra (12th Edition)

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

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 88

Answer

$-\frac{125n^{12}}{r^{6}}$

Work Step by Step

According to the power rule for exponents, $(a^{m})^{n}=a^{mn}$ (where $m$ and $n$ are integers and $a$ is a real number). Therefore, $(\frac{-5n^{4}}{r^{2}})^{3}=(\frac{-1\times5\times n^{4}}{r^{2}})^{3}=\frac{(-1)^{1\times3}5^{1\times3}n^{4\times3}}{r^{2\times3}}=\frac{(-1)^{3}5^{3}n^{12}}{r^{6}}=\frac{-1\times125\times n^{12}}{r^{6}}=\frac{-125n^{12}}{r^{6}}=-\frac{125n^{12}}{r^{6}}$.
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