# How do you simplify (2x^3y^7)^-2?

${\left(2 {x}^{3} {y}^{7}\right)}^{- 2} = \frac{1}{4 {x}^{6} {y}^{14}}$
${x}^{- y} = \frac{1}{x} ^ y$, so.............
${\left(2 {x}^{3} {y}^{7}\right)}^{- 2} = \frac{1}{2 {x}^{3} {y}^{7}} ^ \left(2\right) = \frac{1}{\left(2 {x}^{3} {y}^{7}\right) \left(2 {x}^{3} {y}^{7}\right)}$
$=$ $\frac{1}{{2}^{2} {x}^{6} {y}^{14}} = \frac{1}{4 {x}^{6} {y}^{14}}$