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

Question

UPSC Quizroom · Accepted Answer

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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