Algebra 1: Common Core (15th Edition)

$n=1$ or $n=-1$
Given: $n^{-3}=\frac{1}{n^3}$ and $(\frac{1}{n})^5=\frac{1}{n^5}$ Set two equations equal to each other: $\frac{1}{n^3}=\frac{1}{n^5}$ Solve for $n$: $n^5=n^3$ $n^5-n^3=0$ $n^3(n^2-1)=0$ $n^3=0$ or $n^2-1=0$ $n=0$ or $n=\pm1$ Since $n$ cannot be $0$, the solutions are $n=1$ or $n=-1$.