## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2 - Page 33: 107

#### Answer

$\frac{1}{x^{12}}$

#### Work Step by Step

$$\frac{(x^{-2}y)^{-3}}{(x^{2}y^{-1})^{3}}$$ When a product is raised to a power, raise each factor to that power. $$=\frac{x^{(-2\times-3)}y^{-3}}{x^{(2\times3)}y^{(-1\times3)}}$$ $$=\frac{x^{6}y^{-3}}{x^{(6)}y^{(-3)}}$$ Regroup like terms then simplify $$=(\frac{x^{-6}}{x^{6}})(\frac{y^{-3}}{y^{-3}})$$ Any number divided by itself equals one; and When dividing exponential expressions with the same non-zero base, subtract the denominator's exponent from the numerator's exponent: $$=x^{(-6-6)}\times1$$ $$=x^{-12}$$ When an exponent is negative, write the expression as a fraction and switch the position of the base from the numerator to the denominator (or vise versa) and make the exponent positive. $$=\frac{1}{x^{12}}$$

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