# How do you simplify ((x^-3 y^-4 )/ (x^-5 y^-7 )) ^-3 ?

Mar 3, 2018

See a solution process below:

#### Explanation:

First, use this rule of exponents to divide the $x$ and $y$ terms within the parenthesis:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

${\left(\frac{{x}^{\textcolor{red}{- 3}} {y}^{\textcolor{red}{- 4}}}{{x}^{\textcolor{b l u e}{- 5}} {y}^{\textcolor{b l u e}{- 7}}}\right)}^{-} 3 \implies$

${\left({x}^{\textcolor{red}{- 3} - \textcolor{b l u e}{- 5}} {y}^{\textcolor{red}{- 4} - \textcolor{b l u e}{- 7}}\right)}^{-} 3 \implies$

${\left({x}^{\textcolor{red}{- 3} + \textcolor{b l u e}{5}} {y}^{\textcolor{red}{- 4} + \textcolor{b l u e}{7}}\right)}^{-} 3 \implies$

${\left({x}^{\textcolor{red}{2}} {y}^{\textcolor{red}{3}}\right)}^{\textcolor{b l u e}{- 3}}$

Next, use this rule for exponents to eliminate the outer exponent:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${x}^{\textcolor{red}{2} \times \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{3} \times \textcolor{b l u e}{- 3}} \implies$

${x}^{\textcolor{red}{- 6}} {y}^{\textcolor{red}{- 9}}$

Now, use this rule of exponents to eliminate the negative exponents:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

$\frac{1}{{x}^{\textcolor{red}{- - 6}} {y}^{\textcolor{red}{- - 9}}} \implies$

$\frac{1}{{x}^{\textcolor{red}{6}} {y}^{\textcolor{red}{9}}}$