Factorise :

$3 x^{3}-x^{2}-3 x+1$

Let $f(x)=3 x^{3}-x^{2}-3 x+1$ be the given polynomial. The factors of the constant term $+$ 1 are $\pm 1 .$ The factor of coefficient of $x^{3}$ is $3 .$ Hence, possible rational roots of $f(x)$ are: $\pm \frac{1}{3}$

We have,

$f(1)=3(1)^{3}-(1)^{2}-3(1)+1=3-1-3+1=0$

And $\quad f(-1)=3(-1)^{3}-(-1)^{2}-3(-1)+1=-3-1+3+1=0$

So, $(x-1)$ and $(x+1)$ are factors of $f(x).$

$\Rightarrow(x-1)(x+1)$ is also a factor of $f(x).$

$\Rightarrow x^{2}-1$ is a factor of $f(x).$

Let us now divide $f(x)=3 x^{3}-x^{2}-3 x+1$ by $x^{2}-1$ to get the other factors of $f(x).$

By long division, we have

$\begin{array}{l}x^{2}-1 |\overline {3 x^{3}-x^{2}-3 x+1} (3x-1)\\ \;\;\; \;\;\;\;\;\;\,\; 3 x^{3}\quad\quad\,\,-3x\;\;\;\;\;\;\; \\ \hline \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-x^{2}+1 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-x^{2}+1 \\ \hline \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\; 0 \end{array}$

$\therefore \quad 3 x^{3}-x^{2}-3 x+1=\left(x^{2}-1\right)(3 x-1)$

$\Rightarrow \quad 3 x^{3}-x^{2}-3 x+1=(x-1)(x+1)(3 x-1)$

Hence, $3 x^{3}-x^{2}-3 x+1=(x-1)(x+1)(3 x-1)$