# Simplify and express the final result using positive exponents. (x^{-3}y

Simplify and express the final result using positive exponents.
$$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$

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Aniqa O'Neill
To simplify the expression: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$
Solution:
Given expression is: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$
On simplifying further we get:
$$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({x}^{{-{3}}}\right)}^{{-{2}}}\cdot{\left({y}^{{4}}\right)}^{{-{2}}}$$
$$\displaystyle={\left({x}^{{{\left(-{3}\right)}{\left(-{2}\right)}}}\right)}{\left({y}^{{{4}{\left(-{2}\right)}}}\right)}$$
$$\displaystyle={\left({x}^{{6}}\right)}{\left({y}^{{-{8}}}\right)}$$
$$\displaystyle={\left({x}^{{6}}\right)}{\left({\frac{{{1}}}{{{y}^{{8}}}}}\right)}$$
$$\displaystyle={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$
$$\displaystyle\Rightarrow{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$
Result: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$