If 2s=a+b+c, then what is s(s-a)(s-b)(s-c)\left[\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}-\frac{1}{s}\right] equal to ?

Explanation

Substitute x = s-a, y = s-b, z = s-c, so x+y+z = s. The expression in brackets simplifies to \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z} = \frac{(x+y)(y+z)(z+x)}{xyz(x+y+z)}. Multiplying by sxyz evaluates the whole expression to (x+y)(y+z)(z+x) = c \cdot a \cdot b = abc.

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