Question

若x+y+z=xyz,求证x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz

Answer 1

x(1-y²)(1-z²)+y(1-x²)(1-z²)+z(1-x²)(1-y²)
=x+y+z-x(y²+z²)+xy²z²-y(x²+z²)+yx²z²-z(x²+y²)+zx²y²(xy²z²,yx²z²,zx²y²中分别提取出xyz，用x+y+z代替，并重新组合各项的顺序)
=xyz+(x+y+z)yz-y²z-yz²+(x+y+z)xz-x²z-xz²+(x+y+z)xy-x²y-xy²
=xyz+xyz+xyz+xyz
=4xyz
关键是代换x+y+z=xyz

Answer 2

x(1-y²)(1-z²)+y(1-x²)(1-z²)+z(1-x²)(1-y²)
=x+y+z-x(y²+z²)+xy²z²-y(x²+z²)+yx²z²-z(x²+y²)+zx²y²(xy²z²,yx²z²,zx²y²中分别提取出xyz，用x+y+z代替，并重新组合各项的顺序)
=xyz+(x+y+z)yz-y²z-yz²+(x+y+z)xz-x²z-xz²+(x+y+z)xy-x²y-xy²
=xyz+xyz+xyz+xyz
=4xyz
关键是代换x+y+z=xyz