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

Oct 11, 2015

$\frac{4 {x}^{6}}{9 {y}^{4}}$

#### Explanation:

Try to simply the expression

$\frac{3 {x}^{- 2} {y}^{3}}{2 x y}$

first, then worry about the $- 2$ exponent.

Notice that you can write

$\frac{3 {x}^{- 2} {y}^{3}}{2 x y} = \frac{3 \cdot {y}^{3}}{2 x y} \cdot \frac{1}{x} ^ 2 = \frac{3 {y}^{2}}{2 {x}^{3}}$

The original expression now becomes

${\left(\frac{3 {y}^{2}}{2 {x}^{3}}\right)}^{- 2} = {\left(\frac{2 {x}^{3}}{3 {y}^{2}}\right)}^{2} = \frac{{2}^{2} \cdot {x}^{3 \cdot 2}}{{3}^{2} \cdot {y}^{2 \cdot 2}} = \textcolor{g r e e n}{\frac{4 {x}^{6}}{9 {y}^{4}}}$