8. In $\triangle A B C$, the three sides $a, b, c$ form an arithmetic sequence. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$.
8. From the given condition, $2 b=a+c$, so $2 \sin B=\sin A+\sin C$

Given $A+B+C=\pi$, then $2 \sin (A+C)=\sin A+\sin C$
Therefore, $4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}$
Since $\sin \frac{A+C}{2} \neq 0$, we have $2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2}$
Thus, $5 \cos A-4 \cos A \cos C+5 \cos C=5(\cos A+\cos C)-4 \cos A \cos C$
$$
\begin{array}{l}
=10 \cos \frac{A+C}{2} \cos \frac{A-C}{2}-2[\cos (A+C)+\cos (A-C)] \\
=10 \cos \frac{A+C}{2} \cos \frac{A-C}{2}-2\left[2 \cos ^{2} \frac{A+C}{2}+8 \cos ^{2} \frac{A+C}{2}-2\right] \\
=20 \cos ^{2} \frac{A+C}{2}-20 \cos ^{2} \frac{A+C}{2}+4=4
\end{array}
$$