What Are the Possible Values of r in This Viete Relations Problem?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Relations
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
ehrenfest
Messages
2,001
Reaction score
1
[SOLVED] Viete relations problem

Homework Statement


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]

Homework Equations


http://en.wikipedia.org/wiki/Viète's_formulas

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.
 
Physics news on Phys.org
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?
 
morphism said:
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?

I did mess up. Change yz^2 to y^2 z on the LHS. Anyway I already peeked at the solution.