Given, x + 1⁄x = 4,Find the value of: x^6 + 1⁄x^6 − 3 (x^4 + 1⁄x^4) + 4 (x² + 1⁄x²)

Correct — A. Build each power sum from x + 1⁄x = 4.

Square, subtract 2: x² + 1⁄x² = 4² − 2 = 14
Square again, subtract 2: x⁴ + 1⁄x⁴ = 14² − 2 = 194
Cube identity: x³ + 1⁄x³ = 4³ − 3 × 4 = 52
Square that, subtract 2: x⁶ + 1⁄x⁶ = 52² − 2 = 2702

Substitute: 2702 − 3 × 194 + 4 × 14
Multiply out: 2702 − 582 + 56
Combine: 2176 → option (a).

Why the others are wrong

• (b)
  2087 — 2087 is odd, but every piece of the expression is even: 2702, 582 and 56. Even minus even plus even cannot give an odd number.
• (c)
  2562 — 2562 is 386 too high. It would need the two bracket terms to net −140, but −3 × 194 + 4 × 14 = −582 + 56 = −526.
• (d)
  2476 — 2476 is 300 above the true total. Recheck the pieces: x⁶ + 1⁄x⁶ = 2702, 3 × 194 = 582 and 4 × 14 = 56, which combine to 2176.

Every power sum here comes from one identity. If x + 1⁄x = k, then x² + 1⁄x² = k² − 2, because squaring adds a middle term 2 × x × (1⁄x) = 2.

The same step doubles any power: x⁴ + 1⁄x⁴ = (x² + 1⁄x²)² − 2. For the cube, x³ + 1⁄x³ = k³ − 3k.

x⁶ + 1⁄x⁶ is then the cube sum squared, minus 2. Work out the sums once, then substitute.

Key facts

• If x + 1⁄x = k, then x² + 1⁄x² = k² − 2.
• If x + 1⁄x = k, then x³ + 1⁄x³ = k³ − 3k.
• With k = 4: x² + 1⁄x² = 14, x³ + 1⁄x³ = 52, x⁴ + 1⁄x⁴ = 194 and x⁶ + 1⁄x⁶ = 2702.

Common traps

• Squaring x + 1⁄x = 4 to 16 and forgetting to subtract 2.
• Writing x³ + 1⁄x³ = 4³ = 64 and dropping the −3k term.

No directly related past PYQ was found.