2. If $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}=$
1 , then $a b c=$
2.1 .

From the given, we have
$$
\begin{array}{l}
\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\left(\frac{c}{c a+c+1}-1\right)=0 \\
\Rightarrow \frac{a}{a b+a+1}+\frac{b}{b c+b+1}-\frac{c a+1}{c a+c+1}=0 \\
\Rightarrow\left(\frac{a}{a b+a+1}-\frac{c a}{c a+c+1}\right)+\left(\frac{b}{b c+b+1}-\frac{1}{c a+c+1}\right)=0 \\
\Rightarrow \frac{a(1-a b c)}{(a b+a+1)(c a+c+1)}- \\
\Rightarrow \frac{1-a b c}{(b c+b+1)(c a+c+1)}=0 \\
\Rightarrow \frac{1-a b c}{c a+c+1}\left(\frac{a}{a b+a+1}-\frac{1}{b c+b+1}\right)=0 \\
\Rightarrow-\frac{(a b c-1)^{2}}{(c a+c+1)(a b+a+1)(b c+b+1)}=0 \\
\Rightarrow(a b c-1)^{2}=0 \Rightarrow a b c=1 .
\end{array}
$$