Factorise :

$x^{3}-6 x^{2}+11 x-6$

Let $f(x)=x^{3}-6 x^{2}+11 x+6$ be the given polynomial. The factors of the constant term $-6$ are $±1,±2,±3$ and $±6.$

We have, $\quad f(1)=(1)^{3}-6(1)^{2}+11(1)-6=1-6+11-6=0$

And, $f(2)=(2)^{3}-6(2)^{2}+11(2)-6=8-24+22=6=0$

So, $(x-1)$ and $(x-2)$ are factor of $f(x).$

$\Rightarrow \quad(x-1)(x-2)$ is also factor of $f(x).$

$\Rightarrow \quad x^{3}-3 x+2$ is a factor of $f(x).$

Let us now divide $f(x)=x^{3}-6 x^{2}+11 x-6$ by $x^{2}-3 x+2$ to get the other factors of $f(x).$

By long division, we have

$\begin{array}{l}x ^ { 2 } - 3x + 2 |\overline {x ^ { 3 } - 6 x ^ { 2 } + 1 1 x - 6} (x-3)\\ \;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x^{3}-3 x^{2}+2 x\;\;\;\;\;\;\; \\ \hline \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-3 x^{2}+9 x-6 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,-3 x^{2} + 9 x-6 \\ \hline \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\; 0 \end{array}$

$\therefore x^{3}-6 x^{2}+11 x-6=\left(x^{2}-3 x+2\right)(x-3)$

$\Rightarrow x^{3}-6 x^{2}+11 x-6=(x-1)(x-2)(x-3)$

Hence, $x^{3}-6 x^{2}+11 x-6=(x-1)(x-2)(x-3)$