Question

已知x+y+z=1 x2+y2+z2=2 x3+y3+z3=3 求x4+y4+z4=?

Answer 1

（x+y+z）²-（x²+y²+z²）=2（xy+yz+zx）=-1,xy+yz+zx=-1/2

x3+y3+z3=3xyz+(x+y+z)(x²+y²+z²-xy-yz-zx)=3xyz+1*（2-（-1/2））=3,xyz=1/6

（x2+y2+z2）²=x4+y4+z4-2x²y²-2y²z²-2x²z²=4,故x4+y4+z4=4+2（x²y²+y²z²+x²z²）

x²y²+y²z²+x²z²=（xy+yz+zx）²-2xyz（x+y+z）=（-1/2）²-2*1/6*1=-1/12

x4+y4+z4=4+2*（-1/12）=23/6

不清的话就追问.晚上好!