4. -50 .

If $x \neq 0$, then $\left(x^{2}+\frac{1}{x^{2}}\right)+5\left(x+\frac{1}{x}\right)+6=0$.
Let $m=x+\frac{1}{x}$. Thus,
$$
\begin{array}{l}
m^{2}-2+5 m+6=0 \\
\Rightarrow m_{1}=-1, m_{2}=-4 .
\end{array}
$$

Assume $x_{1} x_{2}$ are the roots of $x+\frac{1}{x}=-1$, i.e., $x^{2}+x+1=0$.
$$
\begin{array}{l}
\text { Hence } x_{1}^{3}+x_{2}^{3} \\
=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=2 .
\end{array}
$$

Similarly, $x_{3}^{3}+x_{4}^{3}=-52$.
Therefore, $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}=-50$.
4. Let the roots of the equation $x^{4}+5 x^{3}+6 x^{2}+5 x+1=0$ in the complex number range be $x_{1}, x_{2}, x_{3}, x_{4}$. Then $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$ $+x_{4}^{3}=$ $\qquad$ .