Compute the sum $S=\sum_{i=0}^{101} \frac{x_{i}^{3}}{1-3 x_{i}+3 x_{i}^{2}}$ for $x_{i}=\frac{i}{101}$.

Answer: $S=51$.

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Since $x_{101-i}=\frac{101-i}{101}=1-\frac{i}{101}=1-x_{i}$ and

$$
1-3 x_{i}+3 x_{i}^{2}=\left(1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}\right)+x_{i}^{3}=\left(1-x_{i}\right)^{3}+x_{i}^{3}=x_{101-i}^{3}+x_{i}^{3},
$$

we have, by replacing $i$ by $101-i$ in the second sum,

$$
2 S=S+S=\sum_{i=0}^{101} \frac{x_{i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}+\sum_{i=0}^{101} \frac{x_{101-i}^{3}}{x_{i}^{3}+x_{101-i}^{3}}=\sum_{i=0}^{101} \frac{x_{i}^{3}+x_{101-i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}=102,
$$

so $S=51$.

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