Question

已知x+y+z=0,x2+y2+z2=1,求xy+yz+xz,x4+y4+z4的解

Answer 1

（x+y+z）^2=[(x+y)+z]^2

=(x^2+2xy+y^2)+z^2+2zx+2zy

=x^2+y^2+z^2+2xy+2xz+2yz

=x^2+y^2+z^2+2(xy+xz+yz)=0

x+y+z=0

xy + xz+yz= -1/2

(xy+xz+yz)^2

=x^2y^2+x^2z^2+y^2z^2+2xzy^2+2yzx^2+2xyz^2

=x^2y^2+x^2z^2+y^2z^2 +2xyz(x+y+z)

=1/4

x^2y^2+x^2z^2+y^2z^2=1/4

(x^2+y^2+z^2)^2

=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2=1

x^4+y^4+z^4= 1/2