3. Solve the inequality

$$
x^{3}-3 x^{2}+x+1+\frac{1}{x-1} \geqslant 0
$$
Solution.

Since $x-1$ is in the denominator, $x \neq 1$.

(2 points)

We solve the inequality by multiplying by the denominator, but we must distinguish two cases depending on the sign of the expression we are multiplying by.

Note. A student who does not pay attention to the sign of the expression when multiplying the inequality can receive a maximum of 6 points.

$1^{\circ} x-1>0:$

Multiplying the inequality by the denominator in this case will not change the inequality sign, so we get

$$
x^{4}-4 x^{3}+4 x^{2} \geqslant 0
$$

which is equivalent to

$$
x^{2}(x-2)^{2} \geqslant 0
$$

The left side of the inequality is a perfect square, so the last inequality is satisfied for all $x \in \mathbb{R}$,

(2 points)

and since we assumed $x>1$, the solution for this case is

$$
x \in\langle 1,+\infty\rangle
$$

(2 points)

$2^{\circ} x-1<0:$

Multiplying the inequality by the denominator in this case changes the inequality sign, so we get

$$
x^{4}-4 x^{3}+4 x^{2} \leqslant 0
$$

which is equivalent to

$$
(*) \quad x^{2}(x-2)^{2} \leqslant 0
$$

The left side of the inequality is a perfect square, so the last inequality is satisfied only if

$$
x^{2}(x-2)^{2}=0
$$

that is, $x=0$ and $x=2$.

But in this case, $x<1$, so the solution for this case is only

$$
x=0 \text { . }
$$

Therefore, the solution to the given inequality is

$$
x \in\{0\} \cup\langle 1,+\infty\rangle
$$

Note. A student who correctly solves the problem but still omits $x=0$ in the second case (because they claim that (*) has no solution) loses 5 points.