jmc

Note that $x^6-3x^4+3x^2-1$ is very similar to $(x-1{)}^3=x^3-3x^2+3x-1$. If we make the substitution $y=x^2$, our expression becomes $x^6-3x^4+3x^2-1=y^3-3y^2+3y-1=(y-1{)}^3$.

Now, we substitute $x^2$ back in for $y$: $(y-1{)}^3=(x^2-1{)}^3$. Note that $x^2-1=(x-1)(x+1)$. Thus, our factorization is $x^6-3x^4+3x^2-1=(x^2-1{)}^3=((x-1)(x+1){)}^3=\boxed{(x-1{)}^3(x+1{)}^3}$.