Question

已知 x+y+z=3 求证(x-1)^3+(y-1)^3+(z-1)^3-3(x-1)(y-1)(z-1)=0

Answer 1

由x+y+z=3

可知（x-1）+（y-1）+（z-1）=0.

令x-1=a

y-1=b

z-1=c

0=a+b+c

c=-(a+b)

(x-1)^3+(y-1)^3+(z-1)^3-3(x-1)(y-1)(z-1)

=a^3+b^3-(a+b)^3-3abc

=a^3+b^3-(a^3+3a^2b+3ab^2+b^3)-3abc

=-3a^2b-3ab^2-3abc

=-3ab(a+b+c)

=-3ab*0

=0