Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set - Page 268: 20

Answer

$\dfrac{1}{-125x^{6}a^{9}} $

Work Step by Step

Using laws of exponents, the given expression simpifies to \begin{align*} & (-5y^0x^2a^3)^{-3} \\\\&= \dfrac{1}{(-5y^0x^2a^3)^{3}} \\\\&= \dfrac{1}{-5^3y^{0(3)}x^{2(3)}a^{3(3)}} \\\\&= \dfrac{1}{-125x^{6}a^{9}} .\end{align*}
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