8. Let the function $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$, where $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ are non-negative integers. Given that $f(1)=4$, $f(5)=152$, then $f(6)=$ $\qquad$
\[
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}+a_{3}=4, \\
a_{0}+5 a_{1}+25 a_{2}+125 a_{3}=152 .
\end{array}\right.
\]
Obviously, \(a_{3}=1\), otherwise 
\[
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}=4, \\
a_{0}+5 a_{1}+25 a_{2}=152
\end{array}\right.
\]
\(\Rightarrow a_{0}+5 a_{1}+25(4-a_{0}-a_{1})=100-24 a_{0}-20 a_{1}=152\), which is a contradiction;
Thus, 
\[
\left\{\begin{array}{l}
a_{0}+a_{1}+a_{2}=3, \\
a_{0}+5 a_{1}+25 a_{2}=27 .
\end{array}\right.
\]
Similarly, \(a_{2}=1, a_{1}=0, a_{0}=2 \Rightarrow f(x)=x^{3}+x^{2}+2\).
Therefore, \(f(6)=216+36+2=254\).