设 a=x-1,b=y-1,c=z-1,于是 a+b+c=0xyz-xy-xz-yz = (a+1)(b+1)(c+1)- (a+1)(b+1)- (b+1)(c+1)- (c+1)(a+1)=abc -2利用a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0得 a^3+b^3+c^3= 3abc,于是所以原式= （3abc-...