解方程组：x+y+z=3； x^2+y^2+z^2=3；x^5+y^5+z^5=3

1）9=(x+y+z)^2= x^2+y^2+z^2+2(xy+xz+yz)=3+2(xy+xz+yz) ==> xy+xz+yz=3
2）9=(x+y+z)(x^2+y^2+z^2)=(x^3+y^3+z^3)+xy(y+x)+xz(x+z)+yz(y+z)
=(x^3+y^3+z^3)+xy(3-z)+xz(3-y)+yz(3-x)
=(x^3+y^3+z^3)+3(xy+xz+yz)-3xyz
=(x^3+y^3+z^3)+9-3xyz
得：x^3+y^3+z^3=3xyz
3）(x^2+y^2+z^2)(x^3+y^3+z^3)=9xyz
(x^5+y^5+z^5)+x^2y^2(x+y)+x^2z^2(x+z)+y^2z^2(y+z)=9xyz
3+x^2y^2(3-z)+x^2z^2(3-y)+y^2z^2(3-x)=9xyz
3+3[(xy+xz+yz)^2-2xyz(x+y+z)]-xyz(xy+xz+yz)=9xyz
3+3[9-6xyz]-3xyz=9xyz
xyz=1
4）因为x+y+z=3, xy+xz+yz=3, xyz=1因此由韦达定理，x,y,z分别为方程P^3-3P^2+3P-1=0的三个根
即(P-1)^3=0
解得P=1为三重根
故x=y=z=1