Question

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

Answer 1

①(x+y+z)^3-(x^3+y^3+z^3)

=[(x+y+z)^3-x^3]-(y^3+z^3)
=(y+z)[(x+y+z)^2+x(x+y+z)+x^2]-(y+z)(y^2-yz+z^2)
=(y+z)[(x+y+z)^2+x(x+y+z)+x^2-y^2+yz-z^2]
=(y+z)(3x^2+3xy+3yz+3zx)
=3(y+z)(x+y)(z+x)
=3(1-x)(1-z)(1-y)
=3(1-x-z+xz)(1-y)
=3(1-y-x+xy-z+yz+xz-xyz)
=3(xy+yz+zx-xyz)
=1^3-3
=-2
即xyz=xy+yz+zx+2/3
②(x+y+z)^2-(x^2+y^2+z^2)
=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2
=2(xy+yz+zx)
=1^2-2
=-1
即xy+yz+zx=-1/2
结合①②，得：xyz=-1/2+2/3=1/6

③(x^2+y^2+z^2)^2-2*(xy+yz+zx)^2
=x^4+y^4+z^4+2(xy)^2+2(yz)^2+2(zx)^2-2(xy)^2-2(yz)^2-2(zx)^2-4xzy^2-4yzx^2-4xyz^2
=(x^4+y^4+z^4)-4xyz(x+y+z)
=(x^4+y^4+z^4)-4*(1/6)*1
=(x^4+y^4+z^4)-2/3
=2^2-2*(-1/2)^2
=4-1/2
=7/2
所以，x^4+y^4+z^4=7/2+2/3=25/6

Answer 2

ok