x+y+z=2

Question

x^2+y^2+z^2=14
x^3+y^3+z^3=20

please help
show workings
thanks in advance

Answer 1

from x+y+z=2
(x+y+z)^2= 4
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 4
using #2:
14 + 2(xy+xz+yz) = 4
xy+xz+yz= -5

(x+y+z)^3 = 8
x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz = 8

20 + 3x^2(y + z) + 3y^2(x + z) + 3z^2(x + y) + 6xyz = 8

starting to get messy
but one solution is
x=1
y=-2
z=3
because of the symmetry, we can arrange these in more ways
e.g.
x = -2
y = 1
z = 3