x+y+z=1
x^(2)+y^(2)+z^(2)=2
x^(3)+y^(3)+z^(3)=3
求:x^(4)+y^(4)+z^(4)=?
1,由(x^2+y^2+z^2)*(x+y+z)=x^3+y^3+z^3+(x+y)z^2+(y+z)x^2+(x+z)y^2,得到(x+y)z^2+(y+z)x^2+(x+z)y^2
=-1
2,由(x+y+z)^2-(x^2+y^2+z^2)=2*(xy+yz+xz)得到xy+yz+xz=-0.5
3,由(x+y+z)^3=x^3+y^3+z^3+6xyz+3*（(x+y)z^2+(y+z)x^2+(x+z)y^2）,得到xyz=1/6
4,由（xy+yz+xz）^2=(x^2*y^2+x^2*z^2+z^2*y^2)+2xyz(x+y+z),得到x^2*y^2+x^2*z^2+z^2*y^2=-1/12
5,由(x^2+y^2+z^2）^2=x^4+y^4+z^4+2*(x^2*y^2+x^2*z^2+z^2*y^2),得到x^4+y^4+z^4=25/6.