If x + 1⁄x = 2√3, then find the value of x³ + 1⁄x³.

Correct — D. Use the cube identity instead of solving for x.

Identity: x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x)
Cube the sum: (2√3)³ = 8 × 3√3 = 24√3
Three times the sum: 3 × 2√3 = 6√3
Subtract: 24√3 − 6√3 = 18√3 → option (d)

Why the others are wrong

• (a)
  10√3 — 10√3 ≈ 17.3 is too small. The roots of x + 1⁄x = 2√3 are √3 ± √2, and the larger, about 3.15, has a cube of about 31 on its own.
• (b)
  6√3 — 6√3 is the correction term 3(x + 1⁄x) = 3 × 2√3 on its own. It is what you subtract from the cube, not the result.
• (c)
  24√3 — 24√3 is (x + 1⁄x)³ alone. Cubing a sum also produces the cross term 3(x + 1⁄x), so 6√3 still has to come off.

Cube both sides of x + 1⁄x = s and expand, using (a + b)³ = a³ + b³ + 3ab(a + b):

(x + 1⁄x)³ = x³ + 1⁄x³ + 3 · x · (1⁄x) · (x + 1⁄x) = x³ + 1⁄x³ + 3s
So x³ + 1⁄x³ = s³ − 3s

The product x · 1⁄x = 1 is what makes these identities work, so x itself is never needed. The square version is x² + 1⁄x² = s² − 2.

Key facts

• If x + 1⁄x = s, then x³ + 1⁄x³ = s³ − 3s.
• If x + 1⁄x = s, then x² + 1⁄x² = s² − 2.
• If x − 1⁄x = d, then x³ − 1⁄x³ = d³ + 3d.
• (√3)³ = 3√3, so (2√3)³ = 8 × 3√3 = 24√3.

Common traps

• Stopping at (2√3)³ = 24√3 and forgetting to subtract 3(x + 1⁄x).
• Cubing 2√3 as 8√3: the √3 is cubed too, and (√3)³ = 3√3.

No directly related past PYQ was found.