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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?

User8976
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Ray Jasson
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3 Answers3

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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

Community
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Surb
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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.

Empy2
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  • Could expand a bit more please..? what do you mean by "final sum"? – Surb Nov 27 '15 at 12:43
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    What do you mean with "999 without 99"? I find your answer very vague, could you elaborate? – Eric S. Nov 27 '15 at 12:54
  • Michael means that, if you expand the expression and gather the $999 \cdot 99$ terms, we have $999 \cdot 99 \cdot ((a-b)+(b-c)+(c-a)) = 0$. Similarly, gathering the terms with $99$, but not with $999$, we have $99(-2c(a-b)-2a(b-c)-2b(c-a))=0$. Do it similarly to the terms with $999$ but not with $99$. They all get canceled out - so it's okay to replace them with $0$. Now it is just a factoring problem. (+1) because it can generalize the problem like the first answer without heavy machinery :) – rkm0959 Nov 27 '15 at 13:37
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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 $$S=-2(a-b)(c-a)(b-c)=-6.$$

Lutz Lehmann
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