# Math Help - solve

1. ## solve

x+y+z=1
x^2+y^2+z^2=21
x^3+y^3+z^3=55

2. Originally Posted by perash
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

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