# How do you simplify  [(4x^-4 y^2)/ (5x^6 y^-3)]^-3 leaving only positive exponents?

May 16, 2016

$\frac{125 {x}^{30}}{64 {y}^{15}}$

#### Explanation:

Given,

${\left(\frac{4 {x}^{-} 4 {y}^{2}}{5 {x}^{6} {y}^{-} 3}\right)}^{\textcolor{b l u e}{- 3}}$

According to the exponent law , ${\left(x y\right)}^{n} = {x}^{n} {y}^{n}$, the $\textcolor{b l u e}{- 3}$ multiplies by each exponent in the numerator and denominator. In addition, the numbers are raised to $\textcolor{b l u e}{- 3}$. Thus,

$= \frac{{4}^{\textcolor{b l u e}{- 3}} {x}^{\left(- 4 \times \textcolor{b l u e}{- 3}\right)} {y}^{\left(2 \times \textcolor{b l u e}{- 3}\right)}}{{5}^{\textcolor{b l u e}{- 3}} {x}^{\left(6 \times \textcolor{b l u e}{- 3}\right)} {y}^{\left(- 3 \times \textcolor{b l u e}{- 3}\right)}}$

Simplifying,

$= \frac{{4}^{-} 3 {x}^{12} {y}^{-} 6}{{5}^{-} 3 {x}^{-} 18 {y}^{9}}$

According to the exponent law , ${x}^{-} \frac{n}{1} = \frac{1}{x} ^ n$. Similarly, $\frac{1}{x} ^ - n = {x}^{n} / 1$. Thus,

$= \frac{{5}^{3} {x}^{12} \cdot 5 {x}^{18}}{{4}^{3} {y}^{9} {y}^{6}}$

Simplifying,

$= \frac{125 {x}^{12} {x}^{18}}{64 {y}^{9} {y}^{6}}$

Using the exponent law , ${x}^{m} {x}^{n} = {x}^{m + n}$, the exponents in the numerator of base $x$ can be added together. Similarly, the exponents in the denominator of base $y$ can be added together.

$= \frac{125 {x}^{12 + 18}}{64 {y}^{9 + 6}}$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{125 {x}^{30}}{64 {y}^{15}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$