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How to solve this ?

  1. Mar 29, 2012 #1
    Given : a + (1/b) = b + (1/c) = c + (1/a)

    prove a=b=c or a+b+c = 0

    I know that a^3 + b^3 + c^3 = 3abc is sufficient proof to derive the above two relations asked for..but I cannot get the expression up to there.

    I tried to see if a+b+c = 0 logically leads to the given equation, but no luck so far.

    What am I missing ?
     
  2. jcsd
  3. Mar 29, 2012 #2
    Assume that the equation is true, and assume that first a solution is a=b=c try replacing a with b:
    b+1/b = b+1/c=c+1/b
    And c should also be equal to b. So replace again:
    c+1/c = c+1/c = c+1/c
    True true true. So indeed that is a solution. Do the same for a+b+c=0 <=> a = -b-c
    and find that the equation is true
     
  4. Mar 29, 2012 #3

    SammyS

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    You have two independent equations:
    a + (1/b) = b + (1/c)​
    and
    a + (1/b) = c + (1/a)
    Solve the second one for c, and plug that into the first one.

    Do some algebraic manipulation. One solution should be a = b .

    The other solution is more complicated. I assume it should lead to a+b+c=0
     
  5. Mar 29, 2012 #4

    tiny-tim

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    hmm … the solution is symmetric, so i'd expect the proof to be symmetric also :redface:

    if we subtract the three equations in pairs, i can get to abc = ±1, or a = b = c :smile:

    and it might help to use that if p/q = r/s = t/u, then they all equal [Ap+Br+Ct]/[Aq+Bs+Cu]
     
  6. Mar 30, 2012 #5
    Putting a+b+c = 0 the equation is not becoming an identity...
     
  7. Mar 30, 2012 #6

    SammyS

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    What else have you tried?
     
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