If x 3 + y 3 = 35 and xy = 6, x > y, then value of (x – y) will be

Let s = x + y and p = xy = 6. Use the identity x³ + y³ = (x + y)³ − 3xy(x + y), so s³ − 18s = 35, which gives s³ − 18s − 35 = 0. Trying s = 5 gives 125 − 90 − 35 = 0, so s = 5. Factoring gives (s − 5)(s² + 5s + 7) = 0, and the quadratic factor has negative discriminant, so s = 5 is the only real root.

Now (x − y)² = (x + y)² − 4xy = 25 − 24 = 1, so x − y = ±1. Since x > y, x − y = 1. Check: x = 3, y = 2 gives 27 + 8 = 35 and xy = 6.

• Option A (1): correct, as shown above.
• Option B (2): would need (x − y)² = 4, so x + y would have to satisfy s² = 28, which does not fit the cubic relation.
• Option C (3): would need (x − y)² = 9, giving s² = 33, which again fails the cubic.
• Option D: wrong, since A is valid.

The core concept is the algebraic identities for sums of cubes and squares, which appear often in Quantitative Aptitude and CSAT reasoning. The method of reducing to a single variable and checking for the only real root is useful in many problems.

Trap: Aspirants often take x − y = ±1 and pick the first option without applying the x > y condition, or they forget to verify that s = 5 is the only real root.