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
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