Evaluate 2012+((a - b)(b - c)(c - a))/(abc)

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The discussion centers on evaluating the expression 2012 + ((a - b)(b - c)(c - a))/(abc) under the condition that (a - b)/c + (b - c)/a + (c - a)/b = 36. Participants confirmed that this condition can be rewritten as -((a - b)(b - c)(c - a))/(abc). The conclusion reached is that the evaluation leads to a specific numeric result based on the relationships between the variables a, b, and c.

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If $a, b, c$ are real numbers with $\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}=36$, evaluate $\dfrac{(a-b)(b-c)(c-a)}{abc}+2012$.
 
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Re: Evaluate 2012+((a-b)(b-c)(c-a))/abc

My solution
It interesting that $\dfrac{a-b}{c} + \dfrac{b-c}{a} + \dfrac{c-a}{b} = -\dfrac{(a-b)(b-c)(c-a)}{abc}$ so adding the constraint to the given expression to be evaluated gives $1976$.
 
Re: Evaluate 2012+((a-b)(b-c)(c-a))/abc

Jester said:
My solution
It interesting that $\dfrac{a-b}{c} + \dfrac{b-c}{a} + \dfrac{c-a}{b} = -\dfrac{(a-b)(b-c)(c-a)}{abc}$ so adding the constraint to the given expression to be evaluated gives $1976$.

Thanks for participating and well done, Jester!

Yes, it took me some time to realize $\dfrac{a-b}{c} + \dfrac{b-c}{a} + \dfrac{c-a}{b} = -\dfrac{(a-b)(b-c)(c-a)}{abc}$! :o
 

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