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Viete relations problem

  • Thread starter ehrenfest
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  • #1
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[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.
 

Answers and Replies

  • #2
morphism
Science Advisor
Homework Helper
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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?
 
  • #3
2,012
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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?
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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