Welcome to our community

Be a part of something great, join today!

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

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
If $a, b, c$ are real numbers with $\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}=36$, evaluate $\dfrac{(a-b)(b-c)(c-a)}{abc}+2012$.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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 $1976$.
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
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 $1976$.
Thanks for participating and well done, Jester!

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