Find $S=(a-b)(99-c)(999-2c)+(b-c)(99-a)(999-2a)+(c-a)(99-b)(999-2b)$

Question

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?

Answer 1

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

Answer 2

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.