If 1/x = 1/p + 1/q, then what is pq/(p² – q²) × ((x+p)/(x–p) – (x+q)/(x–q)) equal to?

From 1/x = 1/p + 1/q, we get x = pq/(p+q). Compute (x+p)/(x–p): x+p = pq/(p+q) + p = (pq + p² + pq)/(p+q) = (p² + 2pq)/(p+q) = p(p+2q)/(p+q). Similarly x–p = (pq – p² – pq)/(p+q) = –p²/(p+q). So (x+p)/(x–p) = (p+2q)/(–p) = –(p+2q)/p. By symmetry (x+q)/(x–q) = –(q+2p)/q. The difference is –(p+2q)/p + (q+2p)/q = [–q(p+2q) + p(q+2p)]/(pq) = [–pq – 2q² + pq + 2p²]/(pq) = 2(p²–q²)/(pq). Multiplying by pq/(p²–q²) gives 2.