College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2: 62

Answer

$=\frac{-27b^{30}}{a^{9}}$

Work Step by Step

$$(\frac{-30a^{14}b^{8}}{10a^{17}b^{-2}})^{3}$$ Group factors with like bases: $$=[(\frac{-30}{10})(\frac{a^{14}}{a^{17}})(\frac{b^{8}}{b^{-2}})]^{3}$$ $$=[-3a^{(14-17)}b^{(8-(-2))}]^{3}$$ $$=[-3a^{-3}b^{(8+2)}]^{3}$$ $$=[-3a^{-3}b^{10}]^{3}$$ When raising a product to a power, raise each factor to that power: $$=(-3)^{3}a^{(-3\times3)}b^{(10\times3)}$$ $$=-27a^{-9}b^{30}$$ Write as a fraction, move the base with the negative exponent to the other side of the fraction, and make the exponent positive: $$=\frac{-27b^{30}}{a^{9}}$$
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