## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2: 61

#### Answer

$\frac{-27b^{15}}{a^{18}}$

#### Work Step by Step

$$(\frac{-15a^{4}b^{2}}{5a^{10}b^{-3}})^{3}$$ Group factors with like bases: $$=[(\frac{-15}{5})(\frac{a^{4}}{a^{10}})(\frac{b^{2}}{b^{-3}})]^{3}$$ $$=[-3a^{(4-10)}b^{(2-(-3))}]^{3}$$ $$=[-3a^{-6}b^{(2+3)}]^{3}$$ $$=[-3a^{-6}b^{5}]^{3}$$ When raising a product to a power, raise each factor to that power: $$=(-3)^{3}a^{(-6\times3)}b^{(5\times3)}$$ $$=-27a^{-18}b^{15}$$ Write as a fraction, move the base with the negative exponent to the other side of the fraction, and make the exponent positive: $$=\frac{-27b^{15}}{a^{18}}$$

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