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Thread: solve

  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
    $\displaystyle x+y+z=1$
    Square both sides,
    $\displaystyle x^2+y^2+z^2 + 2(xy+xz+yz) = 1$
    By second equation we have,
    $\displaystyle 21 + 2(xy+xz+yz) = 1 \implies xy+xz+yz = -10$

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

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

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

    Similar question here.
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