solve conditional identity

Question

if a+b+c=2s,then prove that [1/(s-a)]+[1/(s-b)]+ [1/(s-c)] = abc/s(s-a)(s-b)(s-c)

Answer 1

First let's check whether this is a true identity by using numbers: a=1, b=2, c=5, s=4.

a+b+c=8 and 2s=8 so a+b+c=2s.

1/(s-a)+1/(s-b)+1/(s-c)=⅓+½-1=-⅙;

abc/[s(s-a)(s-b)(s-c)]=10/[(4)(3)(2)(-1)]=-5/12.

Furthermore, if a and/or b and/or c=0, the RHS=0 but the LHS≠0.

This is proof that the given identity is false.