# How do you simplify (\frac { x ^ { - 4} y ^ { 2} } { x ^ { 5} y ^ { 6} } ) ^ { 2}?

##### 1 Answer
Mar 30, 2017

$\left(\frac{1}{{x}^{18} {y}^{8}}\right)$

#### Explanation:

The first thing we should do is multiply each exponent by $2$

${\left(\frac{{x}^{-} 4 {y}^{2}}{{x}^{5} {y}^{6}}\right)}^{2} \to \left(\frac{{x}^{- 4 \cdot 2} {y}^{2 \cdot 2}}{{x}^{5 \cdot 2} {y}^{6 \cdot 2}}\right) \to \left(\frac{{x}^{-} 8 {y}^{4}}{{x}^{10} {y}^{12}}\right)$

What we could do next is rewrite the expression using only positive exponents. You'll notice that ${x}^{-} 8$ is a negative exponent so we have to make positive using this rule: $\left({a}^{- b} = \frac{1}{a} ^ b\right)$ Basically, we are moving the ${x}^{-} 8$ to the denominator.

This would result in:

$\left(\frac{{y}^{4}}{{x}^{8} {x}^{10} {y}^{12}}\right)$

Finally, we can simply the expression using the following rules:

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

${x}^{a} / {x}^{b} = {x}^{a - b}$

Thus:

((y^4)/(x^8x^10y^12)) : x^8*x^10 = x^(8+10) = x^18; y^4/y^12 = 1/(y^8)

So our final answer is...

$\left(\frac{1}{{x}^{18} {y}^{8}}\right)$