## Algebra 1: Common Core (15th Edition)

Recall the equation: $a^{-n}= \frac{1}{a^{n}}$. Thus, we simplify this expression and solve: $a^{3}b^{-1}=\frac{a^{3}}{b}$ We now plug in 2 for a and -4 for b to obtain: $\frac{2^{3}}{-4}=8/-4=-2$