Simplify. [(232)³ + (140)³ + (353)³ − 3 × 232 × 140 × 353] ⁄ [(232)² + (140)² + (353)² − 232 × 140 − 140 × 353 − 353 × 232]

Correct — A. The expression is an image: the numerator is 232³ + 140³ + 353³ − 3 × 232 × 140 × 353, and the denominator is 232² + 140² + 353² − 232×140 − 140×353 − 353×232.

Identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

The denominator is that second bracket, so it cancels with it.

What survives is a + b + c = 232 + 140 + 353 = 725 → option (a)

Why the others are wrong

• (b)
  596 — 596 is not reachable from these three numbers — not their total, and not any of the sums you get by flipping one sign. The cancelled fraction can only leave a + b + c.
• (c)
  445 — 445 is 232 + 353 − 140. The identity adds all three numbers, so no term is subtracted.
• (d)
  261 — 261 is 353 + 140 − 232, the same sign slip with a different term flipped. The bracket left standing is a + b + c, all three added.

The whole item is one factorisation: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

SSC prints the second bracket as the denominator, so the fraction collapses to a + b + c and no cube is ever computed.

Recognition is the skill being tested. The signature to look for is three cubes minus 3abc on top, and three squares minus the three cross products underneath — every cross term negative.

Key facts

• a³ + b³ + c³ − 3abc factors as (a + b + c)(a² + b² + c² − ab − bc − ca).
• So this fraction equals a + b + c whenever its denominator is that second bracket.
• 232 + 140 + 353 = 725.
• A corollary of the same identity: when a + b + c = 0, a³ + b³ + c³ = 3abc.

Common traps

• Starting to cube 232, 140 and 353 by hand
• Missing one of the negative cross terms and deciding the identity does not apply
• Adding only two of the three numbers, which produces 445 or 261

No directly related past PYQ was found.