$\begin{array}{l}\text { 3. Given real numbers } a, b, c. \text { If } \\ \frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 c a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1, \\ \text { then }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2008}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2008}+ \\ \left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2008}=\text {. }\end{array}$
3.3.

Notice
$$
\begin{array}{l} 
\frac{a^{2}-b^{2}-c^{2}}{2 b c}+\frac{b^{2}-c^{2}-a^{2}}{2 a a}+\frac{c^{2}-a^{2}-b^{2}}{2 a b}=-1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+1+\frac{(c-a)^{2}-b^{2}}{2 c a}+1+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}-1=1 \\
\Rightarrow \frac{(b-c)^{2}-a^{2}}{2 b c}+\frac{(c-a)^{2}-b^{2}}{2 c a}+ \\
\frac{(a+b)^{2}-c^{2}}{2 a b}=0 \\
\Rightarrow a\left[(b-c)^{2}-a^{2}\right]+b\left[(c-a)^{2}-b^{2}\right]+ \\
c\left[(a+b)^{2}-c^{2}\right]=0 \\
\Rightarrow a(b-c+a)(b-c-a)+ \\
b(c-a+b)(c-a-b)+ \\
c(a+b+c)(a+b-c)=0 \\
\Rightarrow(a+b-c)\left(a b-a c-a^{2}-b c+a b-\right. \\
\left.b^{2}+a c+b c+c^{2}\right)=0 \\
\Rightarrow(a+b-c)(a+c-b)(b+c-a)=0
\end{array}
$$
$$
\Rightarrow c=a+b \text { or } b=c+a \text { or } a=b+c \text {. }
$$

When $c=a+b$, we have
$$
\begin{array}{l}
\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b-c)^{2}-a^{2}}{2 b c}+1 \\
=\frac{a^{2}-a^{2}}{2 b c}+1=1 .
\end{array}
$$

Similarly, $\frac{c^{2}+a^{2}-b^{2}}{2 c a}=1, \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-1$.
Therefore, the original expression $=3$.
Similarly, when $b=c+a$ or $a=b+c$, the original expression $=3$.