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

问题补充：

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

答案：

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=3

3(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。