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

[anemone]

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

 

[Mathwebster]

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

My solution

Spoiler

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 $1976$.

 

[anemone]

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

My solution

Spoiler

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 $\dfrac{a-b}{c} + \dfrac{b-c}{a} + \dfrac{c-a}{b} = -\dfrac{(a-b)(b-c)(c-a)}{abc}$!