已知x+y+z=0,x2+y2+z2=1,求xy+yz+xz,x4+y4+z4的解

（x+y+z）^2=[(x+y)+z]^2
=(x^2+2xy+y^2)+z^2+2zx+2zy
=x^2+y^2+z^2+2xy+2xz+2yz
=x^2+y^2+z^2+2(xy+xz+yz)=0
x+y+z=0
xy + xz+yz= -1/2
(xy+xz+yz)^2
=x^2y^2+x^2z^2+y^2z^2+2xzy^2+2yzx^2+2xyz^2
=x^2y^2+x^2z^2+y^2z^2 +2xyz(x+y+z)
=1/4
x^2y^2+x^2z^2+y^2z^2=1/4
(x^2+y^2+z^2)^2
=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2=1
x^4+y^4+z^4= 1/2