Given, x + 1⁄x = 4 find the value of x³ + 1⁄x³.

Correct — C. Cube the given sum and take away the cross term.

Identity, since x × 1⁄x = 1: (x + 1⁄x)³ = x³ + 1⁄x³ + 3(x + 1⁄x)
Rearrange: x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x)

Substitute x + 1⁄x = 4: 4³ − 3 × 4
Evaluate: 64 − 12 = 52 → option (c)

Why the others are wrong

• (a)
  55 — 55 = 64 − 9, but the term the identity subtracts is 3 × (x + 1⁄x) = 3 × 4 = 12, leaving 52.
• (b)
  57 — 57 is not a multiple of 4, yet the factor form (x + 1⁄x)(x² − 1 + 1⁄x²) = 4 × 13 shows the answer must be.
• (d)
  50 — 50 is 2 short of 52. Squaring first gives x² + 1⁄x² = 4² − 2 = 14, and then (x + 1⁄x)(x² − 1 + 1⁄x²) = 4 × 13 = 52.

For x + 1⁄x = k, the product of the two terms is 1, so the cube identity (a + b)³ = a³ + b³ + 3ab(a + b) shrinks to x³ + 1⁄x³ = k³ − 3k.

A second route checks it. Square first: x² + 1⁄x² = k² − 2 = 14. Then use a³ + b³ = (a + b)(a² − ab + b²): 4 × (14 − 1) = 52.

Key facts

• If x + 1⁄x = k, then x² + 1⁄x² = k² − 2.
• If x + 1⁄x = k, then x³ + 1⁄x³ = k³ − 3k.
• x⁶ + 1⁄x⁶ = (x³ + 1⁄x³)² − 2, which here is 52² − 2 = 2702.

Common traps

• Writing (x + 1⁄x)³ as x³ + 1⁄x³: cubing a sum adds the cross term 3(x + 1⁄x), worth 12 here.
• Subtracting 3 instead of 3k: the cross term is 3 × x × 1⁄x × (x + 1⁄x), which is 3 × 4 = 12.

No directly related past PYQ was found.