## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2 - Page 33: 112

#### Answer

$\frac{1}{x^{32}y^{40}z^{48}}$

#### Work Step by Step

$$(\frac{x^{4}y^{5}z^{6}}{x^{-4}y^{-5}z^{-6}})^{-4}$$ Within the parentheses, group like terms and simplify. $$=[(\frac{x^{4}}{x^{-4}})(\frac{y^{5}}{y^{-5}})(\frac{z^{6}}{z^{-6}})]^{-4}$$ When dividing exponential expressions with the same non-zero base, subtract the exponent of the denominator from the exponent of the numerator. $$=(x^{(4-(-4))}y^{(5-(-5))}z^{(6-(-6))})^{-4}$$ $$=(x^{(4+4)}y^{(5+5)}z^{(6+6)})^{-4}$$ $$=(x^{8}y^{10}z^{12})^{-4}$$ When a product is raised to an exponent, raise each factor to that exponent. $$=x^{(8\times(-4))}y^{(10\times(-4))}z^{(12\times(-4))}$$ $$=x^{-32}y^{-40}z^{-48}$$ When an exponent is negative, write the expression as a fraction and move the base from the numerator to the denominator (or vise versa) making the exponent positive. $$=\frac{1}{x^{32}y^{40}z^{48}}$$

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