2. Let $a$, $b$, $c$ be integers, and for all real numbers $x$,
$$
(x-a)(x-8)+1=(x-b)(x-c)
$$

always holds. Then the value of $a+b+c$ is $\qquad$.
2.20 or 28 .
$$
\begin{array}{l}
\text { Given } x^{2}-(8+a) x+8 a+1 \\
=x^{2}-(b+c) x+b c
\end{array}
$$

always holds, so,
$$
\left\{\begin{array}{l}
8+a=b+c, \\
8 a+1=b c .
\end{array}\right.
$$

Eliminating $a$ yields $b c-8(b+c)=-63$, which simplifies to $(b-8)(c-8)=1$.
Since $b$ and $c$ are integers, we have $b-8=c-8=1$ or $b-8=c-8=-1$. Thus, $a+b+c=20$ or 28 .