Reducing an algebraic fraction, cyclic in three variables, to another

[brotherbobby]

Homework Statement

If $2s = a+b+c$, prove that $$\boxed{\pmb{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}-\frac{1}{s}=\frac{abc}{s(s-a)(s-b)(s-c)}}}$$

Relevant Equations

I don't know if the following three formulae will be useful, all equivalent to one another and written out in different forms.
1. $ab(a-b)+bc(b-c)+ca(c-a) = -(a-b)(b-c)(c-a)$
2. $a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$
3. $a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)= (a-b)(b-c)(c-a)$

Problem : Let me copy and paste the problem statement as it appears in the text, as shown above.

Attempt : I can sense there is an "elegant" way of doing this, but I don't know how. I show below my attempt using $\text{Autodesk Sketchbook}$. I hope am not violating anything.

Ok so I have got the answer, with clumsy algebra and using brute force.Does someone have hints to an elegant approach?

 

[pasmith]

You can write the numerator as <br /> \begin{split}<br /> (s-a)(s-b)(s-(s-c)) + s(s-c)((s - a) + (s - b)) <br /> &amp;= c(s-a)(s-b) + s(s-c)(2s - a - b) \\<br /> &amp;= c(s-a)(s-b) + sc(s-c) \\<br /> &amp;= c(2s^2 - (a + b + c)s + ab) \\<br /> &amp;= c(2s^2 - 2s^2 + ab) \\<br /> &amp;= abc.<br /> \end{split}