Question 659772
if you have this:

{{{((2xy^-2)/(y^-3*y^3 ))^2}}}, then


{{{((2x(1/y^2))/(1/y^3)*y^3 )^2}}}


{{{((2x/y^2)/(1/cross(y^3))*cross(y^3) )^2}}}


{{{(2x/y^2)^2}}}


{{{4x^2/y^4}}}


but, if you have:


{{{(2xy^-2)^2/(y^-3*y^3 )}}}, then


{{{(2x(1/y^2))^2/(1/y^3*y^3 )}}}


{{{(2x(1/y^2))^2/(1/cross(y^3)*cross(y^3) )}}}


{{{(2x/y^2)^2 )}}}


{{{4x^2/y^4}}}

the solution is same