6. If $x_{t}=\frac{i}{101}$, then $T=\sum_{i=0}^{101} \frac{x_{i}^{3}}{3 x_{t}^{2}-3 x_{2}+1}$ is $(\quad)$.
A. 51
B. 52
C. 53
D. 54
6. A

Because
$$
\begin{aligned}
& \frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{\left(1-x_{i}\right)^{3}}{3\left(1-x_{i}\right)^{2}-3\left(1-x_{i}\right)+1} \\
= & \frac{1-x_{i}^{3}+3 x_{i}^{2}-3 x_{i}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{x_{i}^{3}}{3 x_{i}^{2}-3 x_{i}+1}=1
\end{aligned}
$$

So $2 T=\sum_{i=0}^{101}\left[\frac{x_{j}^{3}}{3 x_{i}^{2}-3 x_{i}+1}+\frac{\left(1-x_{1}\right)^{3}}{\left.3\left(1-x_{i}\right)^{2}-3 i 1-x_{i}\right)+1}\right]$
$$
=102
$$

So $(T=51$, hence the answer is A.)