8 Given the function $f(x)=x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right)$, then for any real numbers $a, b$ $(a+b \neq 0), \frac{f(a)+f(b)}{a^{3}+b^{3}}$ is ( ).
(A) always greater than zero
(B) always equal to zero
(C) always less than zero
(D) the sign is uncertain
8 Because
$$
\begin{aligned}
f(-x)+f(x) & =-x^{3}-\log _{2}\left(\sqrt{x^{2}+1}+x\right)+x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right) \\
& =-\log _{2}\left[\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)\right] \\
& =0,
\end{aligned}
$$

so $f(-x)=-f(x)$, which means $f(x)$ is an odd function.
Furthermore,
$$
\begin{aligned}
f(x) & =x^{3}-\log _{2}\left(\sqrt{x^{2}+1}-x\right) \\
& =x^{3}+\log _{2}\left(\sqrt{x^{2}+1}+x\right)
\end{aligned}
$$

is monotonically increasing on $(0,+\infty)$, so $f(x)$ is an increasing function on $\mathbf{R}$. Noting that $f(a)$ and $f(-b)$ have the same sign as $a-(-b)$, we have $\frac{f(a)+f(b)}{a+b}>0$. Since $a^{3}+b^{3}$ has the same sign as $a+b$, it follows that $\frac{f(a)+f(b)}{a^{3}+b^{3}}>0$. Therefore, the correct choice is A.