Question

已知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,哪位大神帮帮忙啊，详细过程啊

Answer 1

因为
(x^3+y^3+z^3)*(x+y+z)=x^4+y^4+z^4+xy*(x^2+y^2)+xz(x^2+z^2)+yz(y^2+z^2)
=x^4+y^4+z^4+xy(1-z^2)+xz(1-y^2)+yz(1-x^2)
=x^4+y^4+z^4+xy+xz+yz-xyz(x+y+z)
=3
所以
x^4+y^4+z^4=3-(xy+xz+yz)+xyz
因为
x^2+y^2+z^2 =2=2(x+y+z)
x^2-x+y^2-y+z^2-z=x+y+z=1
x(x-1)+y(y-1)+z(z-1)=1
x(y+z)+y(x+z)+z(x+y)=-1
2(xy+xz+yz)=-1
所以
xy+xz+yz=-1/2
因为
x^3+y^3+z^3=3=3(x+y+z)
x^3-x+y^3-y+z^3-z=2(x+y+z)=2
所以
x^2(1-x)+y^2(1-y)+z^2(1-z)=x^2(y+z)+y^2(x+z)+z^2(x+y)=xy(x+y)+xz(x+z)+yz(y+z)=xy(1-z)+xz(1-y)+yz(1-x)=xy+xz+yz-3xyz=-1/2-3xyz=2
xyz=-5/6
所以
x^4+y^4+z^4=3-(-1/2)-5/6=8/3

Answer 2

不会…