设x+y+z=3,求代数式(3(xyz-xy-xz-yz)+6)/((x-1)^3+(y-1)^3+

设x+y+z=3,求代数式(3(xyz-xy-xz-yz)+6)/((x-1)^3+(y-1)^3+

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则 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======以下答案可供参考======供参考答案1:x+y+z=3， (3(xyz-xy-xz-yz)+6)/((x-1)³+(y-1)³+(z-1)³)=1========计算得有点复杂了，应该有简单方法。xyz-xy-xz-yz=xyz-xy-xz-yz+x+y+z-1 -x-y-z+1=xy(z-1)-x(z-1)-y(z-1)+(z-1) -x-y-z+1=(z-1)(xy-x-y+1) -x-y-z+1=(z-1)(x(y-1)-(y-1)) -x-y-z+1=(z-1)(y-1)(x--1) -x-y-z+1=(z-1)(y-1)(x--1) -2 ……x+y+z=33(xyz-xy-xz-yz)+6=3(x-1)(y-1)(z-1)(x-1)³+(y-1)³+(z-1)³ …… a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)=(x-1)³+(y-1)³+(z-1)³-3(x-1)(y-1)(z-1)+ 3(x-1)(y-1)(z-1)=(x-1+y-1+z-1)((x-1)²+(y-1)²+(z-1)²-(x-1)(y-1)-(x-1)(z-1)-(y-1)(z-1))+3(x-1)(y-1)(z-1)=3(x-1)(y-1)(z-1)=====令 x-1=a, y-1=b, z-1=c， 代入原式计算更方便，步骤略。

这个解释是对的