If p²+q²−29 = 2pq−20 = 52−2pq, then the difference between the maximum and minimum possible value of (p³−q³) is:

Options

  1. 486
  2. 189
  3. 378
  4. 243

Correct Answer

378

Explanation

Solution: Step 1: Solve for pq 2pq − 20 = 52 − 2pq → 4pq = 72 → pq = 18 Step 2: Solve for p² + q² p² + q² − 29 = 2pq − 20 → p² + q² − 29 = 2×18 − 20 → p² + q² − 29 = 16 → p² + q² = 45 Step 3: Use formula for p³ − q³ p³ − q³ = (p − q)(p² + pq + q²) p² + pq + q² = 45 + 18 = 63 (p − q)² = p² − 2pq + q² = 45 − 36 = 9 → p − q = ±3 Then p³ − q³ = 63 × 3 = 189 or 63 × (−3) = −189 Step 4: Difference between maximum and minimum Difference = 189 − (−189) = 378 Answer: 378


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