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Math Help - find x^100 +y^100 +z^100

  1. #1
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    find x^100 +y^100 +z^100

    x,y,z real numbers

    x+y+z=3
    x^2+y^2+z^2=25
    x^4+y^4 +z^4=209
    find
    x^100 +y^100 +z^100
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  2. #2
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    Where did you get it from?

    We have,
    x+y+z=3 ...................(1)
    x^2+y^2+z^2=25......(2)
    x^4+y^4+z^4=209....(3)

    We shall use the identity,
    (a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc) (*)

    Square equation (1) to get by (*),
    (x^2+y^2+z^2)+2(xy+xz+yz)=9
    Substitute (2) to get,
    25+2(xy+xz+yz)=9
    Thus,
    xy+xz+yz=-8..............(4)

    Square equation (2) to get by (*),
    (x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)=625
    Substitute (3) to get,
    209+2(x^2y^2+x^2z^2+y^2z^2)=625
    Thus,
    x^2y^2+x^2z^2+y^2z^2=208...(5)

    Square equation (4) to get by (*),
    (x^2y^2+x^2z^2+y^2z^2)+2(x^2yz+xy^2z+xyz^2)=64
    Substitute (4) to get,
    208+2(x^2yz+xy^2z+xyz^2)=64
    208+2xyz(x+y+z)=64
    Substitute (1) to get,
    208+6xyz=64
    Thus,
    xyz=-24

    Thus, we have 3 equations,
    x+y+z=3
    xy+xz+yz=-8
    xyz=-24

    To solve these equations we shall use Viete's theorem.

    That means the solutions to the polynomial,
    t^3-3t^2-8t+24=0
    Are the solutions of those three equations.
    Factor,
    t^2(t-3)-8(t-3)=0
    (t-3)(t^2-8)=0
    Thus,
    t=3 and t=sqrt(8) and t=-sqrt(8)

    Thus, one solution without lose of generality is,
    x=3
    y=sqrt(8)
    z=-sqrt(8)

    Thus,
    x^100+y^100+z^100=3^100+8^50+8^50=1.42724769270595 98810582859694495e+95
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Where did you get it from?

    ...3^100+8^50+8^50=1.42724769270595988105828596944 95e+95
    Hello, TPH,

    I'm just curious where you got this number from?

    When I used my calculator the result is:

    3^100+8^50+8^50 = 518232016117423250798577701704520262974873015249 = 5.1823.. 10^47

    EB
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello, TPH,

    I'm just curious where you got this number from?
    I am one of those calculating prodigies.
    However, I realized my superior skills fails when numbers get too large. Thus, perhaps I made a mistake.
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