## Intermediate Algebra (12th Edition)

a. C. $\frac{1}{64}$ b. D. $-\frac{1}{64}$ c. D. $-\frac{1}{64}$ d. C. $\frac{1}{64}$
According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$. Therefore, a. $4^{-3}=\frac{1}{4^{3}}=\frac{1}{4\times4\times4}=\frac{1}{64}$ b. $-4^{-3}=-\frac{1}{4^{3}}=-\frac{1}{4\times4\times4}=-\frac{1}{64}$ c. $(-4)^{-3}=\frac{1}{(-4)^{3}}=\frac{1}{-4\times-4\times-4}=\frac{1}{-64}=-\frac{1}{64}$ d. $-(-4)^{-3}=-\frac{1}{(-4)^{3}}=-\frac{1}{-4\times-4\times-4}=-\frac{1}{-64}=-(-\frac{1}{64})=\frac{1}{64}$