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

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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
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

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
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