Question

已知xyz满足z+y+z=xyz 求证：

Answer 1

x(1-y²)(1-z²)+y(1-x²)(1-z²)+z(1-x²)(1-y²)

=x(1-y²-z²+y²z²)+y(1-x²-z²+x²z²)+z(1-x²-y²+x²y²)

=x-xy²-xz²+xy²z²+y-yx²-yz²+yx²z²+z-zx²-zy²+zx²y²

=x+y+z-x²y-xy²-x²z-xz²-y²z-yz²+(x+y+z)yz+(x+y+z)xz+(x+y+z)xy

=xyz+xyz+xyz+xyz

=4xyz