2. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a-b+c=7, \\
a b+b c+b+c^{2}+16=0 .
\end{array}
$$

Then the value of $\left(a^{-1}-b^{-1}\right)^{a k c}(a+b+c)^{a+b+c}$ is $\qquad$ .
2. -1 .

From equation (1), we get
$$
(-b)+(a+c+1)=8 \text {. }
$$

From equation (2), we get
$$
(a+c+1)(-b)=c^{2}+16 \text {. }
$$

Therefore, $a+c+1$ and $-b$ are the two roots of the equation
$$
x^{2}-8 x+c^{2}+16=0
$$

Thus, we have $\Delta=(-8)^{2}-4\left(c^{2}+16\right) \geqslant 0$.
Hence, $c^{2} \leqslant 0$.
It is easy to see that $c=0$. Consequently, $x_{1}=x_{2}=4$, which means $a+c+1=4$, $-b=4$, or equivalently, $a=3, b=-4$.
$$
\begin{array}{l}
\text { Therefore, }\left(a^{-1}-b^{-1}\right)^{2 k}(a+b+c)^{a+b+c} \\
=\left(3^{-1}+4^{-1}\right)^{0}[3+(-4)+0]^{3-4+0}=-1 \text {. }
\end{array}
$$