# Simplify ((x^-1y^4)/(x^2y^7))^-3 ?

Feb 21, 2017

${\left(\frac{{x}^{-} 1 {y}^{4}}{{x}^{2} {y}^{7}}\right)}^{-} 3 = {x}^{9} {y}^{9}$

#### Explanation:

The expression as written in the (original - before restoration) question is ambiguous as it is unclear whether denominator should be read as the product of the ${x}^{2}$ and ${y}^{7}$ terms or only the ${x}^{2}$.

I have chosen to interpret the expression as ${\left(\frac{{x}^{-} 1 {y}^{4}}{{x}^{2} {y}^{7}}\right)}^{-} 3$

Two rules of indices will apply here:

(i) ${a}^{m} \times {a}^{n} = {a}^{m + n}$
(ii) ${\left({a}^{m}\right)}^{n} = {a}^{m \times n}$

${\left(\frac{{x}^{-} 1 {y}^{4}}{{x}^{2} {y}^{7}}\right)}^{-} 3 = {\left({x}^{- 1 - 2} \times {y}^{4 - 7}\right)}^{-} 3$ [Rule (i)]

$= {\left({x}^{-} 3 \times {y}^{-} 3\right)}^{-} 3$

$= {x}^{9} {y}^{9}$ [Rule (ii)]