4. It is known that the equation $x^{3}-x-1=0$ has a unique real root $x_{0}$. Come up with at least one equation of the form

$$
a \cdot z^{3}+b \cdot z^{2}+c \cdot z+d=0
$$

where $a, b, c, d$ are integers and $a \neq 0$, one of the roots of which is the number

$$
z=x_{0}^{2}+3 \cdot x_{0}+1
$$
Solution: Let's write down the relations

$$
\begin{gathered}
z=x_{0}^{2}+3 \cdot x_{0}+1 \\
z \cdot x_{0}=x_{0}^{3}+3 \cdot x_{0}^{2}+x_{0} \\
z \cdot x_{0}^{2}=x_{0}^{4}+3 \cdot x_{0}^{3}+x_{0}^{2}
\end{gathered}
$$

The right-hand sides can be simplified (reduced modulo $x_{0}^{3}-x_{0}-1$), using the fact that $x_{0}^{3}=x_{0}+1$. As a result, we get

$$
\begin{gathered}
z=x_{0}^{2}+3 \cdot x_{0}+1 \\
z \cdot x_{0}=3 \cdot x_{0}^{2}+2 \cdot x_{0}+1 \\
z \cdot x_{0}^{2}=2 \cdot x_{0}^{2}+4 \cdot x_{0}+3
\end{gathered}
$$

The first two equations can be considered as a system of linear equations with two unknowns $x_{0}$ and $x_{0}^{2}$. Solving this system, we find $x_{0}=\frac{3 z-2}{z+7}, x_{0}^{2}=\frac{z^{2}-3 z-1}{z+7}$. Substituting these relations into the last equation, we obtain the desired equation in terms of $z$.

Answer: For example, $z^{3}-5 z^{2}-10 z-11=0$.