2.168. $\frac{x^{4}-x^{3}-x+1}{x^{3}-5 x^{2}+7 x-3} \cdot|x-3|$.

2.168. $\frac{x^{4}-x^{3}-x+1}{x^{3}-5 x^{2}+7 x-3} \cdot|x-3|$. 

(Note: The mathematical expression is the same in both languages, so the translation is identical to the original text.)
Solution.

DZ: $\left\{\begin{array}{l}x \neq 1 \\ x \neq 3\end{array}\right.$

$$
\begin{aligned}
& \frac{x^{4}-x^{3}-x+1}{x^{3}-5 x^{2}+7 x-3} \cdot|x-3|=\frac{(x-1)^{2}\left(x^{2}+x+1\right)}{(x-1)^{2}(x-3)} \cdot|x-3|= \\
& =\frac{\left(x^{2}+x+1\right) \cdot|x-3|}{x-3}=\left\{\begin{array}{l}
\frac{\left(x^{2}+x+1\right)(x-3)}{x-3}=-\left(x^{2}+x+1\right), \text { if } 1 \neq x3
\end{array}\right.
\end{aligned}
$$

Answer: $-\left(x^{2}+x+1\right)$ when $x \in(-\infty ; 1) \cup(1 ; 3) ; x^{2}+x+1$ when $x \in(3 ; \infty)$.