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- Feb 14, 2012
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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 $\dfrac{(a-b)(b-c)(c-a)}{abc}+2012$.
Thanks for participating and well done, Jester!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 $1976$.