Answer:[tex]\displaystyle A=\frac{ 4 }{ x^{14} y^{8}}[/tex]Step-by-step explanation:Properties of PowersThe algebraic expression called power has the form x^y where x is called the base and y is called the exponent.Powers have some properties, some of which we'll recall below[tex](x^a)^b=x^{ab}[/tex][tex](x^a)(x^b)=x^{a+b}[/tex][tex]\displaystyle \frac{x^a}{x^b}=x^{a-b}[/tex][tex]\displaystyle x^{-a}=\frac{1}{x^a}[/tex][tex]\displaystyle x^{a}=\frac{1}{x^{-a}}[/tex]We'll use those properties to simplify the expression[tex]\displaystyle A=\left( \frac{ (2)^{-3}(x^{-3}) (y^2) }{(4^{-2}) (x^4) (y^6)} \right)^2[/tex]Taking the power 2 of every term[tex]\displaystyle A=\frac{ (2)^{-6}(x^{-6}) (y^4) }{(4^{-4}) (x^8) (y^{12})}[/tex]Moving the terms with negative exponent to its opposite side[tex]\displaystyle A=\frac{ (2)^{-6} (4^{4})(y^4) }{ (x^8)(x^{6}) (y^{12})}[/tex]Operating the explicit same bases[tex]\displaystyle A=\frac{ (2)^{-6} (4^{4}) }{ (x^{14}) (y^{12-4})}[/tex]Since [tex]4^4=(2^2)^4=2^8[/tex][tex]\displaystyle A=\frac{ (2)^{-6} (2^8) }{ (x^{14}) (y^{12-4})}[/tex]Operating the remaining like bases subtractions[tex]\displaystyle A=\frac{ (2^2) }{ (x^{14}) (y^{8})}[/tex]Finally[tex]\displaystyle A=\frac{ 4 }{ x^{14} y^{8}}[/tex]