For all values for which the fraction is defined, how do you simplify #(a^-1-b^-1)/(b^-2-a^-2)#?

1 Answer
Jan 7, 2017

Answer:

#(a^(-1)-b^(-1))/(b^(-2)-a^(-2)) = -(ab)/(a+b)#

Explanation:

Multiply both numerator and denominator by #a^2b^2# and simplify as follows:

#(a^(-1)-b^(-1))/(b^(-2)-a^(-2)) = (a^2b^2(a^(-1)-b^(-1)))/(a^2b^2(b^(-2)-a^(-2)))#

#color(white)((a^(-1)-b^(-1))/(b^(-2)-a^(-2))) = (a^2b^2a^(-1)-a^2b^2b^(-1))/(a^2b^2b^(-2)-a^2b^2a^(-2))#

#color(white)((a^(-1)-b^(-1))/(b^(-2)-a^(-2))) = (ab^2-a^2b)/(a^2-b^2)#

#color(white)((a^(-1)-b^(-1))/(b^(-2)-a^(-2))) = (ab(b-a))/((a-b)(a+b))#

#color(white)((a^(-1)-b^(-1))/(b^(-2)-a^(-2))) = -(abcolor(red)(cancel(color(black)((a-b)))))/(color(red)(cancel(color(black)((a-b))))(a+b))#

#color(white)((a^(-1)-b^(-1))/(b^(-2)-a^(-2))) = -(ab)/(a+b)#