Problem 1

Compute the sum

S = \sum_{i=0}^{101} \frac{x_i^3}{1 - 3x_i + 3x_i^2}

for

x_i = \frac{i}{101}

Step 2 of 4: Rewrite the denominator as a symmetric sum of cubes

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In plain words

The coefficients $1, -3, 3$ are a classic binomial signature of $(1 - x)^3$ ; completing the cube exposes the hidden symmetry between $x_i$ and $1 - x_i$ .

1 - 3x_i + 3x_i^2 = (1 - 3x_i + 3x_i^2 - x_i^3) + x_i^3 = (1 - x_i)^3 + x_i^3 = x_{101-i}^3 + x_i^3

Detailed analysis

Notice that $1 - 3x_i + 3x_i^2$ contains the first three terms of the binomial expansion $(1 - x_i)^3 = 1 - 3x_i + 3x_i^2 - x_i^3$ . Subtracting and adding $x_i^3$ gives $1 - 3x_i + 3x_i^2 = (1 - x_i)^3 + x_i^3 = x_{101-i}^3 + x_i^3$ , so the general summand becomes $\frac{x_i^3}{x_{101-i}^3 + x_i^3}$ .