8. Given a function $y=f(x)$ defined on $\mathbf{R}$ that satisfies: $f(x)+f(-x)=0$, and when $x \leqslant 0$, we have
$$
y=x^{2 n+1}+x^{2 n}+\cdots+x \text {. }
$$

Then the expression for $f(x)$ on $\mathbf{R}$ is ( ).
(A) $x\left(|x|^{2 n}+|x|^{2 n-1}+\cdots+1\right)$
(B) $|x|\left(x^{2 n}+x^{2 n-1}+\cdots+1\right)$
(C) $x\left[|x|^{2 n}-|x|^{2 n-1}+\cdots+\right.$
$$
\left.(-1)^{i}|x|^{2 n-i}+\cdots+1\right]
$$
(D) $|x|\left[x^{2 n}-x^{2 n-1}+\cdots+\right.$
$$
\left.(-1)^{i} x^{2 n-i}+\cdots+1\right]
$$
8.C.

Since $f(x)=-f(-x)$, i.e., $y=f(x)$ is an odd function, thus, if when $x \leqslant 0$, we have $y=\varphi(x)$, then
$$
y=f(x)=\left\{\begin{array}{ll}
-\varphi(-x), & x>0 ; \\
\varphi(x), & x \leqslant 0 .
\end{array}\right.
$$

When $x \neq 0$, this piecewise expression can be uniformly represented as
$$
y=f(x)=-\frac{x}{|x|} \varphi(-|x|) \text {. }
$$
$$
\begin{array}{l}
\text { Hence } y=-\frac{x}{|x|}\left[(-|x|)^{2 n+1}+(-|x|)^{2 n}+\right. \\
\cdots+(-|x|)] \\
=x\left[|x|^{2 n}-|x|^{2 n-1}+\cdots+(-1)^{i}|x|^{2 n-i}+\cdots+1\right] \text {. } \\
\end{array}
$$

When $x=0$, this expression still holds.