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  1. #1
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    solve

    x+y+z=1
    x^2+y^2+z^2=21
    x^3+y^3+z^3=55
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  2. #2
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    Quote Originally Posted by perash View Post
    x+y+z=1
    x^2+y^2+z^2=21
    x^3+y^3+z^3=55
    x+y+z=1
    Square both sides,
    x^2+y^2+z^2 + 2(xy+xz+yz) = 1
    By second equation we have,
    21 + 2(xy+xz+yz) = 1 \implies xy+xz+yz = -10

    Now, use the identity,
    x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - xz - yz) = (1)(21 - (-10)) = 31
    So,
    x^3+y^3+z^3 - 3xyz = 31
    Use equation 3,
    55 - 3xyz = 31
    Thus,
    xyz=8

    Thus, we have,
    x+y+z=1
    xy+yz+xz=-10
    xyz=8

    The solution to this are the roots of the polynomial, by Viete's Theorem.
    t^3+t -10t+8t=0
    Which solves nicely

    Similar question here.
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