If (a + 1⁄a) = 7√3, then what is the value of (a⁶ + a⁻⁶)?

Correct — C. Climb one power at a time: reach a³ + 1⁄a³ first, then square it.

Identity: x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x)
= (7√3)³ − 3 × 7√3

(7√3)³ = 343 × 3√3 = 1029√3
1029√3 − 21√3 = 1008√3

Now square, since (a³ + 1⁄a³)² = a⁶ + 1⁄a⁶ + 2:
(1008√3)² = 1016064 × 3 = 3048192
a⁶ + a⁻⁶ = 3048192 − 2 = 3048190 → option (c)

Why the others are wrong

• (a)
  3048625 — 3048625 is 145³ exactly. With a² + 1⁄a² = 145, the cube identity still owes its −3(a² + 1⁄a²) term: 145³ − 435 = 3048190. Dropping that correction is the whole trap.
• (b)
  3048542 — 3048542 fits neither route — it is 83 below 145³ and 350 above (1008√3)². The only corrections in play are −435 and −2, and neither lands on a number ending 42.
• (d)
  3048132 — 3048132 sits 58 below the answer and 493 below 145³. Squaring removes exactly 2, so any route reaching this value has already slipped before the final subtraction.

Powers of a + 1⁄a climb a fixed ladder, and the item asks for a rung on it rather than for a itself:

x² + 1⁄x² = (x + 1⁄x)² − 2
x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x)

The sixth power is reached fastest by squaring the cube, because (a³ + 1⁄a³)² = a⁶ + 1⁄a⁶ + 2.

That −2 turns up every time you square a sum of reciprocal powers: the cross term is 2 × a³ × 1⁄a³ = 2, and it has to come back off.

Key facts

• x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x).
• x² + 1⁄x² = (x + 1⁄x)² − 2, so 7√3 gives 147 − 2 = 145.
• (7√3)³ = 1029√3, and 1029√3 − 21√3 = 1008√3.
• 1008² × 3 = 3048192, so a⁶ + a⁻⁶ = 3048190.

Common traps

• Cubing a² + 1⁄a² and forgetting the −3(a² + 1⁄a²) correction, which lands on 3048625.
• Squaring 1008√3 as 1008² and losing the factor 3.
• Adding 2 rather than subtracting it at the final step.

No directly related past PYQ was found.