2. Given that $a$, $b$, and $c$ are real numbers, and
$$
\sqrt[3]{(a+2008)^{8}+8}+\sqrt{|b+8|+1}=\sqrt{6 c-c^{2}} \text {. }
$$

Then the value of $\left(b+c^{2}\right)^{a}$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 4
2.C.

It is easy to know that $(a+2008)^{8}+8 \geqslant 8$,
$$
|b+8|+1 \geqslant 1 \text {. }
$$

Then $\sqrt[3]{(a+2008)^{8}+8}+\sqrt{|b+8|+1} \geqslant \sqrt[3]{8}+\sqrt{1}=3$
$$
\Rightarrow \sqrt{6 c-c^{2}} \geqslant 3 \Rightarrow 6 c-c^{2} \geqslant 9 \Rightarrow(c-3)^{2} \leqslant 0 \text {. }
$$

Since $(c-3)^{2} \geqslant 0$, thus, $(c-3)^{2}=0$, which means $c=3$.
Therefore, the equalities in (1) and (2) must hold, i.e.,
$$
\sqrt[3]{(a+2008)^{8}+8}=8, \sqrt{|b+8|+1}=1 \text {. }
$$

Solving these, we get $a=-2008, b=-8$.
Thus, $\left(b+c^{2}\right)^{a}=\left(-8+3^{2}\right)^{-2008}=1$.