If 2s = a + b + c, then what is s(s−a)(s−b)(s−c)[1/(s−a) + 1/(s−b) + 1/(s−c) − 1/s] equal to?

Note that s−a = (b+c−a)/2, s−b = (a+c−b)/2, s−c = (a+b−c)/2. Also (s−a)+(s−b)+(s−c) = 3s−(a+b+c) = 3s−2s = s. The expression inside brackets is [(s(s−b)(s−c)+s(s−a)(s−c)+s(s−a)(s−b)−(s−a)(s−b)(s−c)]/[s(s−a)(s−b)(s−c)]. After simplification using the identity, the entire expression evaluates to abc.