x+y+z=3,x+y+z-3=0,(x-1)+(y-1)+(z-1)=0
令 a=x-1,b=y-1,c=z-1
則 x=a+1,y=b+1,z=c+1,a+b+c=0
(3(xyz-xy-xz-yz)+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3(xyz-xy-xz+x-yz+y-x-y)+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3(xy(z-1)-x(z-1)-y(z-1)-(x+y))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3((z-1)(xy-x-y)-(x+y))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3((z-1)(xy-x-y+1-1)-(x+y))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3((z-1)(x(y-1)-(y-1)-1)-(x+y))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3((z-1)((y-1)(x-1)-1)-(x+y))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3((z-1)(y-1)(x-1)-z+1-x-y)+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3(z-1)(y-1)(x-1)+1-(x+y+z))+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3(z-1)(y-1)(x-1)+1-3)+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=(3(z-1)(y-1)(x-1)-6+6)/((x-1)^3+(y-1)^3+(z-1)^3)
=3abc/(a^3+b^3+c^3)
=3abc/[(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc] … a³+b³+c³=(a+b+c)(a²+b²+c²-ab-bc-ac)+3abc
=3abc/[0*(a^2+b^2+c^2-ab-bc-ac)+3abc]
=3abc/(3abc)
=1