# How do you simplify ((2^-1x^-3y^-1)(4x^-2)^0)/(2x^-4y^-3)^2?

Mar 18, 2018

${\left(x y\right)}^{5} / 8$

#### Explanation:

$\frac{\left({2}^{-} 1 {x}^{-} 3 {y}^{-} 1\right) {\left(4 {x}^{-} 2\right)}^{0}}{2 {x}^{-} 4 {y}^{-} 3} ^ 2$

We should first notice the power of $0$. Anything to the power of $0$ returns $1$.

$\frac{\left({2}^{-} 1 {x}^{-} 3 {y}^{-} 1\right) \cancel{{\left(4 {x}^{-} 2\right)}^{0}}}{2 {x}^{-} 4 {y}^{-} 3} ^ 2$

$\implies \frac{{2}^{-} 1 {x}^{-} 3 {y}^{-} 1}{2 {x}^{-} 4 {y}^{-} 3} ^ 2$

Now, let's square all of the terms in the bottom:

$\frac{{2}^{-} 1 {x}^{-} 3 {y}^{-} 1}{2 {x}^{-} 4 {y}^{-} 3} ^ 2$

$\implies \frac{{2}^{-} 1 {x}^{-} 3 {y}^{-} 1}{{2}^{2} {x}^{-} 8 {y}^{-} 6}$

Now we compare terms and combine like-terms:

$\frac{\textcolor{b l u e}{{2}^{-} 1} \textcolor{red}{{x}^{-} 3} \textcolor{\mathmr{and} a n \ge}{{y}^{-} 1}}{\textcolor{b l u e}{{2}^{2}} \textcolor{red}{{x}^{-} 8} \textcolor{\mathmr{and} a n \ge}{{y}^{-} 6}}$

$\implies \frac{\textcolor{red}{{x}^{5}} \textcolor{\mathmr{and} a n \ge}{{y}^{5}}}{\textcolor{b l u e}{{2}^{3}}}$

We can finally simplify the way it is written:

$\frac{{x}^{5} {y}^{5}}{{2}^{3}}$

$\implies \textcolor{g r e e n}{{\left(x y\right)}^{5} / 8}$