$(x+1)^3-(x-1)^3-6(x-1)(x+1)\\=[(x+1)^3-(x-1)^3]-6(x^2-1^2)\\=[(x+1)-(x-1)][(x+1)^2+(x+1)(x-1)+(x-1)^2]-6(x^2-1)\\=(x+1-x+1)[(x^2+2x+1)+(x^2-1^2)+(x^2-2x+1)]-6x^2+6\\=2[x^2+2x+1+x^2-1+x^2-2x+1]-6x^2+6\\=2(3x^2+1)-6x^2+6\\=6x^2+2-6x^2+6\\=8$
Vậy $(x+1)^3-(x-1)^3-6(x-1)(x+1)=8$