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Viete relations problem

  1. May 21, 2008 #1
    [SOLVED] Viete relations problem

    1. The problem statement, all variables and given/known data
    Find all real numbers r for which there is at least one triple (x,y,z) of nonzero real numbers such that

    [tex] x^2 y + yz^2 + z^2 x = xy^2 + yz^2 + zx^2 = rxyz[/tex]


    2. Relevant equations
    http://en.wikipedia.org/wiki/Viète's_formulas


    3. The attempt at a solution
    This is equivalent to finding the possible values of r+s+t = 1/r + 1/s + 1/t where r,s,t are real but I don't see how that leads to a solution.


    Fix r and assume that x,y,z exist. Let f(t) = t^3 + at^2 + bt+c be the monic polynomial with
    x,y,z as its zeros. By assumption c is not zero. Its not hard to show that ab = (3+2r)c and a^3 = x^3+ y^3+z^3 + (3+2r)c using Viete's relations. But I am not sure what to do with those or how to get any sort of condition on r.

    Please just provide a hint.
     
  2. jcsd
  3. May 22, 2008 #2

    morphism

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    Just to be sure, did you type out the equations correctly? Or is it supposed to be [itex]\sum x^2 y = rxyz[/itex] instead?
     
  4. May 22, 2008 #3
    I did mess up. Change yz^2 to y^2 z on the LHS. Anyway I already peeked at the solution.
     
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