som

Merk op dat $$\begin{aligned} S & = \sum_{i=0}^{101} \left(\frac{x_i^3}{1-3x_i+3x_i^2} \equiv \frac{x_i^3}{(1-x_i)^3+x_i^3} \equiv \frac{x_i^3}{x_{101-i}^3+x_i^3}\right) \\ & = \sum_{i=0}^{50} \left(\frac{x_i^3}{x_{101-i}^3+x_i^3} + \frac{x_{101-i}^3}{x_{101-i}^3+x_i^3}\right) = \sum_{i=0}^{50} 1 = 51.\end{aligned}$$