已知x+y+z=0,x^2+y^2+z^2=4,求x^4+y^4+z^4的值。

x+y+z=0
(x+y+z)^2=0
x^2+y^2+z^2+2xy+2xz+2yz=0
2(xy+xz+yz)=-4
xy+xz+yz=-2
x^2+y^2+z^2=4
(x^2+y^2+z^2)^2=16=x^4+y^4+z^4+2(x^2y^2+x^2z^2+y^2z^2)
(xy+xz+yz)^2=x^2y^2+x^2z^2+y^2z^2+2(x^2yz+xy^2z+xyz^2)
x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)=4
x^2y^2+x^2z^2+y^2z^2=4
16=x^4+y^4+z^4+2(x^2y^2+x^2z^2+y^2z^2)
x^4+y^4+z^4=16-2(x^2y^2+x^2z^2+y^2z^2)=16-8=8
x^4+y^4+z^4=8

你是不是忘了加这个条件(x+y+z)^3=1

由题意(x+y+z)^3=1

x^3+y^3+z^3+2x^2(y+z)+2z^2(x+y)+2y^2(x+z)+3xyz=1

整理得xyz=0

不妨设x=0

又xy+yz+xz=0.5[(x+y+z)^2-(x^2+y^2+z^2)]=-0.5
则y+z=1 y^3+z^3=1

y^4+z^4=(y+z)(y^3+z^3)-yz(y^2+z^2)=4

故x^4+y^4+z^4=4