jmc

We can factor $x^9-x^6+x^3-1$ as $$x^6(x^3-1)+(x^3-1)=(x^6+1)(x^3-1)=(x^6+1)(x-1)(x^2+x+1).$$Thus, $x^9-x^6+x^3-1$ is a multiple of $x^2+x+1,$ so the remainder is $\boxed{0}.$