The positive numbers p and q satisfy two conditions: p² + q² = 7, and p³q + pq³ = 7. What is the value of p³ + q³?
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Answer & Explanation
Correct answer
A
The two givens aren't in standard form, so rewrite them using the sum s = p + q and the product m = pq.
The second condition, p³q + pq³, factors as pq(p² + q²), which equals m × 7. Since that equals 7, we get 7m = 7, so m = pq = 1. This factoring is the load-bearing move; the surface form hides it.
The first condition says p² + q² = 7. Using (p + q)² = p² + q² + 2pq, we get s² = 7 + 2(1) = 9. Since p and q are positive, s = p + q = 3.
The sum of the cubes is s³ − 3ms = 3³ − 3(1)(3) = 27 − 9 = 18. So the answer is A.
Note that p and q individually are the irrational roots of t² − 3t + 1 = 0, namely (3 + √5)/2 and (3 − √5)/2. Because only the combined quantities s and m are pinned down, plugging in an answer choice gives no shortcut; verifying any choice requires the full derivation.
The distractors all come from mishandling the correction term in the sum-of-cubes expansion. 27 (D) drops the entire correction and reports just (p + q)³. 24 (C) keeps the formula's shape but drops the coefficient 3. 21 (B) uses coefficient 2 instead of 3. 36 (E) flips the sign and adds 9 instead of subtracting it. Each is a genuine partial-work path, not random filler.