# How do you simplify and write ((x^2)^-3(x^-2) )/ (x^2)^-4 with positive exponents?

Apr 5, 2016

$1$

#### Explanation:

Given: $\frac{\textcolor{b r o w n}{{\left({x}^{2}\right)}^{- 3}} \textcolor{red}{\left({x}^{-} 2\right)}}{\textcolor{b l u e}{{\left({x}^{2}\right)}^{- 4}}}$

$\textcolor{b r o w n}{{\left({x}^{2}\right)}^{-} 3} = {x}^{-} 6 = \frac{1}{x} ^ 6$

$\textcolor{red}{\left({x}^{-} 2\right)} = \frac{1}{x} ^ 2$

$\frac{1}{\textcolor{b l u e}{{\left({x}^{2}\right)}^{-} 4}} = \frac{1}{x} ^ - 8 = {x}^{8}$

$\frac{\textcolor{b r o w n}{{\left({x}^{2}\right)}^{- 3}} \textcolor{red}{\left({x}^{-} 2\right)}}{\textcolor{b l u e}{{\left({x}^{2}\right)}^{- 4}}} = \textcolor{b r o w n}{{\left({x}^{2}\right)}^{-} 3} \times \textcolor{red}{\left({x}^{-} 2\right)} \times \frac{1}{\textcolor{b l u e}{{\left({x}^{2}\right)}^{-} 4}}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = \frac{1}{x} ^ 6 \times \frac{1}{x} ^ 2 \times {x}^{8} / 1$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = {x}^{8} / {x}^{8}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = 1$