jmc

Since $x^2+x+1$ is a factor of $(x^2+x+1)(x-1)=x^3-1,$ and so also a factor of $x(x^3-1)=x^4-x,$ the remainder when $(x^4-1)(x^2-1)$ is divided by $x^2+x+1$ is the same as the remainder of $$(x-1)(x^2-1)=x^3-x^2-x+1.$$This has the same remainder when $1-x^2-x+1=-x^2-x+2=-(x^2+x+1)+3$ is divided by $x^2+x+1,$ which is $\boxed{3}.$