Explanation

This is the classic partial-fractions identity Σ a²/((a − b)(a − c)) = 1. Considering the rational function f(t) = t²/((t − a)(t − b)(t − c)) and its partial-fraction decomposition, the sum of the coefficients of the simple-pole terms equals the coefficient of 1/t in the expansion at infinity, which is 1; hence p + q + r = 1 identically (for distinct a, b, c). Squaring gives (p + q + r)² = 1. Hence option (c). The distractors 9, 4, 0 would correspond to sums of 3, 2, 0, which fail for any numeric test, e.g. a = 1, b = 2, c = 3 gives p + q + r = 1.