If p3 – q3 = (p – q) {(p + q)2 – x p q} then the value

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a³ − b³ = (a – b) (a² + ab + b²)
= (a – b) {(a + b)² – ab}
On comparing with
p³ – q³ = (p – q) {(p + q)² – x pq)}, x
= 1