Do the grouping $3x^{3}-x^{2}-3x+1=\left(3x^{3}-x^{2}\right)+\left(-3x+1\right)$, and factor out $x^{2}$ in the first and $-1$ in the second group.
$$x^{2}\left(3x-1\right)-\left(3x-1\right)$$
Factor out common term $3x-1$ by using distributive property.
$$\left(3x-1\right)\left(x^{2}-1\right)$$
Consider $x^{2}-1$. Rewrite $x^{2}-1$ as $x^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
