8. Given the cubic equation
$$
x^{3}-x^{2}-5 x-1=0
$$

has three distinct roots $x_{1}, x_{2}, x_{3}$. Then
$$
\begin{array}{l}
\left(x_{1}^{2}-4 x_{1} x_{2}+x_{2}^{2}\right)\left(x_{2}^{2}-4 x_{2} x_{3}+x_{3}^{2}\right)\left(x_{3}^{2}-4 x_{3} x_{1}+x_{1}^{2}\right) \\
=
\end{array}
$$
8. 444 .

Given the equation factorized as
$$
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=0 \text {. }
$$

then $x_{1}+x_{2}+x_{3}=1$,
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-5, x_{1} x_{2} x_{3}=1 . \\
\text { Also } x_{1}^{3}=x_{1}^{2}+5 x_{1}+1, x_{2}^{3}=x_{2}^{2}+5 x_{2}+1, \\
x_{3}^{3}=x_{3}^{2}+5 x_{3}+1,
\end{array}
$$

we get $x_{1}^{2}-4 x_{1} x_{2}+x_{2}^{2}=\frac{x_{1}^{3}-x_{2}^{3}}{x_{1}-x_{2}}-5 x_{1} x_{2}$
$$
\begin{array}{l}
=\frac{\left(x_{1}^{2}+5 x_{1}+1\right)-\left(x_{2}^{2}+5 x_{2}+1\right)}{x_{1}-x_{2}}-5 x_{1} x_{2} \\
=x_{1}+x_{2}+5-5 x_{1} x_{2}=1-x_{3}+5-\frac{5}{x_{3}} \\
=-\frac{1}{x_{3}}\left(x_{3}^{2}-6 x_{3}+5\right) \\
=-\frac{1}{x_{3}}\left(x_{3}-1\right)\left(x_{3}-5\right) .
\end{array}
$$

Let $f(x)=x^{3}-x^{2}-5 x-1$
$$
=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right) \text {. }
$$

Therefore, $\left(x_{1}^{2}-4 x_{1} x_{2}+x_{2}^{2}\right)\left(x_{2}^{2}-4 x_{2} x_{3}+x_{3}^{2}\right)$.
$$
\begin{aligned}
& \left(x_{3}^{2}-4 x_{3} x_{1}+x_{1}^{2}\right) \\
= & -\frac{1}{x_{1} x_{2} x_{3}} f(1) f(5) \\
= & -f(1) f(5)=444 .
\end{aligned}
$$