Answer
$(x-3)(x+1)(x-1) $
Work Step by Step
We factor by grouping as follows
$$
x^3 -3x^2 - x +3= x^2(x-3)-(x-3)
.$$
Now, factoring $x-3$ out, we have
$$
x^2(x-3)-(x-3) =(x-3)(x^2-1)
.$$
Since $(x^2-1)$ is a difference of two squares, we have
$$
(x-3)(x^2-1)=(x-3)(x+1)(x-1)
.$$
where we used the formula $(x-y)(x+y)=x^2-y^2$.
