If a+b+c = 2s, prove that a/(s-a) +b/(s-b) +c/(s-c) +2 = abc/(s-a)(s-b)(s-c)

Question

If a,b & c are real numbers and (s-a) is not equal to 0, (s-b) is not equal to 0, (s-c) is not equal to 0,  & a+b+c = 2s prove that a/(s-a) +b/(s-b) +c/(s-c) + 2 = abc/(s-a)(s-b)(s-c)

Answer 1

a+b+c=2s, so c=2s-a-b and ac=2as-a^2-ab, bc=2bs-ab-b^2

2=2(s-a)(s-b)(s-c)/((s-a)(s-b)(s-c))=

(2(s-a)(s^2-s(b+c)+bc)/((s-a)(s-b)(s-c))=

2(s^3-s^2(a+b+c)+s(bc+ac+ab)-abc)/((s-a)(s-b)(s-c)).

Combine all the fractions. Just the numerators, because the denominator (s-a)(s-b)(s-c) is common:

a(s-b)(s-c)+b(s-a)(s-c)+c(s-a)(s-b)+2(s^3-s^2(a+b+c)+s(bc+ac+ab)-abc)=

as^2-a(b+c)s+abc + bs^2-b(a+c)+abc + cs^2-c(a+b)s+abc + 2(s^3-s^2(a+b+c)+s(bc+ac+ab)-abc=

2s^3+ (a+b+c)s^2 - 2(ab+ac+bc)s + 3abc

       -2(a+b+c)s^2 + 2(ab+ac+bc)s - 2abc=

2s^3 -(a+b+c)s^2                                abc = 2s^3-2s^3+abc, because a+b+c=2s.

That leaves abc as the numerator so the result is abc/((s-a)(s-b)(s-c)) QED