Given, x − 1⁄x = 3, find the value of x^4 + 1⁄x^4.

Correct — A. Square twice, adjusting for the constant cross term each time.

Square: (x − 1⁄x)² = x² − 2 + 1⁄x²
So 3² = 9 = x² + 1⁄x² − 2
Add 2: x² + 1⁄x² = 11

Square again: (x² + 1⁄x²)² = x⁴ + 2 + 1⁄x⁴
So 11² = 121 = x⁴ + 1⁄x⁴ + 2
Subtract 2: x⁴ + 1⁄x⁴ = 119 → option (a)

Why the others are wrong

• (b)
  125 — 125 would need x² + 1⁄x² = √127 ≈ 11.27. Squaring x − 1⁄x = 3 fixes it at exactly 11, which leads to 119.
• (c)
  130 — 130 would need x² + 1⁄x² = √132 ≈ 11.49, but x − 1⁄x = 3 gives 9 + 2 = 11 exactly.
• (d)
  145 — 145 would need x² + 1⁄x² = √147 ≈ 12.12. From x − 1⁄x = 3 it is 11, and 11² − 2 = 119.

Expressions in x and 1⁄x are climbed by squaring. The product x × 1⁄x = 1 turns every cross term into a plain number.

(x − 1⁄x)² = x² + 1⁄x² − 2 and (x + 1⁄x)² = x² + 1⁄x² + 2. Squaring x² + 1⁄x² always brings a + 2, so x⁴ + 1⁄x⁴ = (x² + 1⁄x²)² − 2.

Key facts

• If x − 1⁄x = k, then x² + 1⁄x² = k² + 2.
• If x + 1⁄x = k, then x² + 1⁄x² = k² − 2.
• x⁴ + 1⁄x⁴ = (x² + 1⁄x²)² − 2.
• If x + 1⁄x = k, then x³ + 1⁄x³ = k³ − 3k.

Common traps

• Using the plus-sign rule at the first step: 9 − 2 = 7 belongs to x + 1⁄x = 3 and leads to 47.
• Adding 2 at the second step to get 123: squaring x² + 1⁄x² always adds 2, so it must be taken away.

No directly related past PYQ was found.