Given: x + 1⁄x = −2, then determine the value of x⁶ + 1⁄x⁶ − 3 (x⁴ + 1⁄x⁴) + 4 (x² + 1⁄x²).

Correct — D.
Clear the fraction: x² + 1 = −2x
Rearrange: x² + 2x + 1 = 0, that is (x + 1)² = 0
So x = −1

Every power in the expression is even: x² = x⁴ = x⁶ = 1
So each bracket = 1 + 1 = 2

Substitute: 2 − 3(2) + 4(2)
= 2 − 6 + 8 = 4 → option (d)

Why the others are wrong

• (a)
  5 — 5 would need x⁶ + 1⁄x⁶ = 3, since 3 − 6 + 8 = 5. With x = −1 that bracket is 1 + 1 = 2.
• (b)
  8 — 8 is the last term alone, 4 × 2. The first two terms add 2 − 6 = −4, which brings the total down to 4.
• (c)
  3 — 3 would need x⁶ + 1⁄x⁶ = 1, since 1 − 6 + 8 = 3. That counts x⁶ = 1 but drops 1⁄x⁶, which is also 1.

x + 1⁄x = ±2 pins x down. For real x, x + 1⁄x is never strictly between −2 and 2. It equals −2 only at x = −1, and 2 only at x = 1.

Once x = −1, every even power is 1, so every bracket of the form xⁿ + 1⁄xⁿ with n even equals 2.

The identity route agrees: x² + 1⁄x² = (−2)² − 2 = 2, and x⁶ + 1⁄x⁶ = 2³ − 3 × 2 = 2.

Key facts

• x + 1⁄x = −2 forces x = −1, because x² + 2x + 1 = (x + 1)² = 0
• x² + 1⁄x² = (x + 1⁄x)² − 2
• x⁶ + 1⁄x⁶ = (x² + 1⁄x²)³ − 3(x² + 1⁄x²)
• For real x, x + 1⁄x ≥ 2 when x > 0 and x + 1⁄x ≤ −2 when x < 0

Common traps

• Using x² + 1⁄x² = (x + 1⁄x)² + 2, which gives 6 instead of 2
• Expanding every identity from scratch when x = −1 already sets each bracket at 2

No directly related past PYQ was found.