Find $S=(a-b)(99-c)(999-2c)+(b-c)(99-a)(999-2a)+(c-a)(99-b)(999-2b)$ Given $(a-b)(b-c)(c-a)=3$
Find $$S=(a-b)(99-c)(999-2c)+(b-c)(99-a)(999-2a)+(c-a)(99-b)(999-2b)$$
Any formula or link related to equation like this?
 A: I will show that 

$$(a−b)(t−c)(s−kc)+(b−c)(t−a)(s−ka)+(c−a)(t−b)(s−kb)=-k(a−b)(b−c)(c−a)$$
  for all $a,b,c,s,t,k\in\Bbb R$.   

Let $s,t,k\in\Bbb R$ and 
$$f(a,b,c)= (a−b)(t−c)(s−kc)+(b−c)(t−a)(s−ka)+(c−a)(t−b)(s−kb)+k(a−b)(b−c)(c−a).$$
We have 
$$\frac{\partial^2}{\partial a^2}f(a,b,c)=k(b-c)-k(b-c)=0$$
This implies that we have
$$\frac{\partial}{\partial a}f(a,b,c)=\alpha$$
For some constant $\alpha$. Now, note that
$$\frac{\partial}{\partial a}f(a,b,c)\Big|_{a=b=c=0}=ts-ts=0$$
and thus $\alpha=0$. Hence $f$ is constant in $a$. Doing the same for $b$ and $c$, it follows that $f$ is constant, i.e. there exists $\delta$ such that $f(a,b,c)=\delta$ for all $a,b,c$. Note that $f(0,0,0)=0$ and thus $\delta = 0$. This shows the claim. 
Set $t=99,s=999$ and $k=2$ to get your solution.
Why k for multiplying in (a-b) (b-c) (c-a),  it could be any of those t or s
A: How many times 999*99?
How many times 999 without 99?
How many time 99 without 999?
Since they don't occur in the final sum, replace them with zero.
A: For any polynomial $p$ of degree $2$, and using divided difference notation as in Newton interpolation, one gets
\begin{align}
(a-b)\,p(c)+(b-c)\,p(a)+(c-a)\,p(b)&=(a-b)(p(c)-p(a))+(c-a)(p(b)-p(a))
\\
&=(a-b)(c-a)\,p[c,a]+(c-a)(b-a)\,p[b,a]
\\
&=-(a-b)(c-a)(b-c)\,p[a,b,c]
\end{align}
Now the twice divided difference $p[a,b,c]$ is a polynomial of degree zero, that is, a constant. It is also equal to the second derivative of $p$ at some intermediary point, thus twice the leading coefficient of $p$. 
In the given situation, $p(x)=(x-r_1)(x-r_2)$, that leading coefficient is $1$, so the expression has the value $$S=-2(a-b)(c-a)(b-c)=-6.$$
