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