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Linear simultaneous equation

  1. Apr 22, 2012 #1
    Can any solve this x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
  2. jcsd
  3. Apr 22, 2012 #2


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    Plan A: multiply out (x+y+z)^2 and (x+y+z)^3.
    From that and the given eqations, you can get the values of yz + zx + xy and xyz.
    Then, you can write down a cubic equation whose roots are x y and z.

    Plan B: take a guess that the solution will probably be integers, and find 3 integers whose squares sum to 35. If you don't get lucky, try plan A :smile:
  4. Apr 22, 2012 #3
    5, -3 & -1 work.... now about there order?
  5. Apr 29, 2012 #4
    You are correctbut why don't you show your working?
  6. Apr 29, 2012 #5
    x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97

    Honestly, I just "guess and checked".


    There isn't a way to find the order as far as I can see.
  7. Apr 29, 2012 #6


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    Why do you think order would matter here? And, why is this posted in "Mechanical Engineering"?
  8. Apr 30, 2012 #7
    Am sorry that this is post in the wrong position but this is the solution x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=35 substitute x+y+z=1 we have (1)^2-2(xy+yz+zx)=35 therefore (xy+yz+zx)=-17
    also x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+zx)+3(xyz)=97
    therefore substitute we have
    (1)^3-3(1)(-17)+3(xyz)=97 which give (xyz)=15 now factor of 15 whose sum is 1 give 5,-1,-3
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