# How do you simplify (2r^-4s^2)^-3 and write it using only positive exponents?

Nov 15, 2017

${r}^{12} / \left(8 {s}^{6}\right)$

#### Explanation:

${x}^{-} a = \frac{1}{x} ^ a$ and vice versa.

Therefore: ${r}^{-} 4 = \frac{1}{r} ^ 4$ and ${\left(2 {r}^{-} 4 {s}^{2}\right)}^{-} 3 = \frac{1}{2 {r}^{-} 4 {s}^{2}} ^ 3$

${\left(2 {r}^{-} 4 {s}^{2}\right)}^{-} 3 = \frac{1}{2 {r}^{-} 4 {s}^{2}} ^ 3 = \frac{1}{2 \left(\frac{1}{r} ^ 4\right) {s}^{2}} ^ 3$

which is:

$= \frac{1}{\frac{2 {s}^{2}}{r} ^ 4} ^ 3$

This is = $\frac{1}{\frac{8 {s}^{6}}{r} ^ 12}$ which is = ${r}^{12} / \left(8 {s}^{6}\right)$