# How do you simplify ((16x^2y^-1)/(4x^0y^-4z))^-2?

Apr 14, 2017

$\frac{16 {z}^{2}}{256 {x}^{4} {y}^{8}}$

#### Explanation:

Recall: ${\left(\frac{a}{b}\right)}^{-} 2 = {\left(\frac{b}{a}\right)}^{2}$

(Flip the fraction and the index changes sign.)

${\left(\frac{16 {x}^{2} {y}^{-} 1}{4 {x}^{0} {y}^{-} 4 z}\right)}^{-} 2 = {\left(\frac{4 {x}^{0} {y}^{-} 4 z}{16 {x}^{2} {y}^{-} 1}\right)}^{2}$

Now simplify the fraction inside the bracket.

${\left(\frac{4 {x}^{0} {y}^{-} 4 z}{16 {x}^{2} {y}^{-} 1}\right)}^{2} = {\left(\frac{4 \left(1\right) y z}{16 {x}^{2} {y}^{4}}\right)}^{2}$

${\left(\frac{4 z}{16 {x}^{2} {y}^{3}}\right)}^{2} \text{ } \leftarrow$ now square all the factors.

$= \frac{16 {z}^{2}}{256 {x}^{4} {y}^{8}}$