设xyz属于R,x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=设xyz属于R,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的值

Question

设xyz属于R,x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=
设xyz属于R,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的值

Answer 1

把x+y+z=1两边平方可以得出
xy+xz+yz=-1/2
再把x^2+y^2+z^2=2两边平方
（x^3+y^3)+(x^3+z^3)+(y^3+z^3)=6;
展开化简可先求出xyz的值
再xy+xz+zyz=-1/2平方,可以得
x^2y^2+x^2z^2+y^2z^2=-1/12
再x^2+y^2+z^2=2平方
细心的你一定可以算出25/6