Answer
$-\frac{125n^{12}}{r^{6}}$
Work Step by Step
According to the power rule for exponents, $(a^{m})^{n}=a^{mn}$ (where $m$ and $n$ are integers and $a$ is a real number).
Therefore, $(\frac{-5n^{4}}{r^{2}})^{3}=(\frac{-1\times5\times n^{4}}{r^{2}})^{3}=\frac{(-1)^{1\times3}5^{1\times3}n^{4\times3}}{r^{2\times3}}=\frac{(-1)^{3}5^{3}n^{12}}{r^{6}}=\frac{-1\times125\times n^{12}}{r^{6}}=\frac{-125n^{12}}{r^{6}}=-\frac{125n^{12}}{r^{6}}$.