x+y+z=6 can be rewritten:
2-x+2-y+2-z=0, so let a=2-x, b=2-y, c=2-z then: a+b+c=0 and c=-(a+b).
The given expression can be written:
a3+b3-(a+b)3+3ab(a+b)=
a3+b3-(a3+3a2b+3ab2+b3)+3ab(a+b)=
-3a2b-3ab2+3a2b+3ab2=0.
Therefore, back substituting a, b and c to regain the original expression:
(2-x)3+(2-y)3+(2-z)3-3(2-x)(2-y)(2-z)=0.