Question 976839
{{{(-a^2b^-2)/(a^-2b^3/a^4b^-2)}}}


{{{(-a^2(1/b^2))/((1/a^2)b^3/a^4(1/b^2))}}}


{{{(-a^2/b^2)/((b^3/a^2)/(a^4/b^2))}}}


{{{(-a^2/b^2)/((b^3b^2)/(a^2*a^4))}}}


{{{(-a^2/b^2)/(b^5/a^6)}}}


{{{(-a^2*a^6)/(b^2*b^5)}}}


{{{-a^8/b^7}}}  

or {{{-a^8*b^-7}}} 

so, answer is C {{{-a^8b^-7}}}