MBA数学题 请帮忙解一道题

等式两边同乘以x(1－yz)·y(1－xz)
得：(x²－yz)·y(1－xz)=(y²－xz)·x(1－yz）
→x²y-x³yz-y²z+xy²z²=xy²-xy³z-x²z+x²yz²
移项后：x²y-xy²+x²z-y²z=x²yz²-xy³z+x³yz-xy²z²
→xy(x-y)+z(x²-y²)=xyz(xz-y²+x²-yz)
→(x-y)[xy+z(x+y)]=xyz[z(x-y)+(x+y)(x-y)]
→(x-y)(xy+xz+yz)=xyz(x-y)(x+y+z)
→(xy+xz+yz)=xyz(x+y+z)
两边同除以xyz得：1/x+1/y+1/z=x+y+z

证明：(x²－yz)/x(1－yz)=(y²－xz)/y(1－xz)
x²y-y²z-x3yz+xy2z2-xy2+x2z+xy3z-x2yz2=0
xy(x-y)+z(x+y)(x-y)+xyz(yz-x2+y2-xz)=0
xy(x-y)+z(x+y)(x-y)-xyz(x-y)(x+y+z)=0
(x-y)[xy+z-xyz(x+y+z)]=0
x不等于y
xy+z-xyz(x+y+z)=0
x,y,z不等于0
左右两边同时除以xyz得x+y+z=1/x+1/y+1/z