Shortcut Trick If p + q + r = 1, pqr = 1, and pq + qr + rp = 1, then p, q, and r are the roots of the cubic equation t^(3) − t^(2) + t − 1 = 0. Factorizing the equation: (t − 1)(t^(2) + 1) = 0, which gives the roots 1, i, and −i. Sum of cubes = p^(3) + q^(3) + r^(3) = (1)^(3) + (i)^(3) + (−i)^(3). Simplifying: 1 − i + i = 1. ∴ The correct answer is 1. Alternate Method Given: p + q + r = 1 pqr = 1 1/p + 1/q + 1/r = 1 Formula Used: p^(3) + q^(3) + r^(3) − 3pqr = (p + q + r)(p^(2) + q^(2) + r^(2) − (pq + qr + rp)) p^(2) + q^(2) + r^(2) = (p + q + r)^(2) − 2(pq + qr + rp) Calculations: ⇒ From 1/p + 1/q + 1/r = 1, we get (pq + qr + rp) / pqr = 1 ⇒ Since pqr = 1, it follows that pq + qr + rp = 1 ⇒ Now, substitute these into the cubic identity: ⇒ p^(3) + q^(3) + r^(3) − 3pqr = (p + q + r)[(p + q + r)^(2) − 3(pq + qr + rp)] ⇒ p^(3) + q^(3) + r^(3) − 3(1) = 1 × [1^(2) − 3(1)] ⇒ p^(3) + q^(3) + r^(3) − 3 = 1 − 3 ⇒ p^(3) + q^(3) + r^(3) − 3 = −2 ⇒ p^(3) + q^(3) + r^(3) = 3 − 2 = 1 ∴ The correct answer is 1. Related Concepts: Standard Cubic Identity The expression a^(3) + b^(3) + c^(3) − 3abc is always equal to (a + b + c)(a^(2) + b^(2) + c^(2) − ab − bc − ca). Special Case: Sum is Zero If a + b + c = 0, then the identity simplifies to a^(3) + b^(3) + c^(3) = 3abc. Vieta's Formulas for Cubic Equations For a cubic equation ax^(3) + bx^(2) + cx + d = 0, the sum of roots is −b/a, the sum of roots taken two at a time is c/a, and the product of roots is −d/a.