If x > 1 and x² + 1⁄x² = 83, then x³ − 1⁄x³ is:

Correct — C. The bridge is (x − 1⁄x)² = x² + 1⁄x² − 2, which converts the given 83 into the difference the question needs.

(x − 1⁄x)² = 83 − 2 = 81
x − 1⁄x = 9, taking the positive root because x > 1

x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x)
= 9³ + 3 × 9
= 729 + 27 = 756 → option (c)

The same value by the other route: x³ − 1⁄x³ = (x − 1⁄x)(x² + 1 + 1⁄x²) = 9 × 84 = 756.

Why the others are wrong

• (a)
  876 — The answer must be 9 × (x² + 1 + 1⁄x²) = 9 × 84, so it is a multiple of 9. 876 ÷ 9 = 97.3…, which rules it out before any cubing is done.
• (b)
  884 — 884 ÷ 9 = 98.2…, so it fails the same multiple-of-9 test. Once x − 1⁄x = 9 is fixed, every value the expression can take is 9 times a whole number.
• (d)
  754 — 754 ÷ 9 = 83.7…, so it too is not a multiple of 9. Using 83 instead of 84 as the second factor gives 747, which the paper does not offer either.

This is the x ± 1⁄x family, and the whole family runs on one observation: x and 1⁄x multiply to 1, so every cross term is a plain number.

That gives the two squares (x − 1⁄x)² = x² + 1⁄x² − 2 and (x + 1⁄x)² = x² + 1⁄x² + 2, which are how you move between the second power and the first.

For the cube, a³ − b³ = (a − b)³ + 3ab(a − b) with ab = 1 collapses to x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x). Nothing here needs the value of x itself, and finding it — solving x² − 9x − 1 = 0 — is slower and messier.

Key facts

• (x − 1/x)² = x² + 1/x² − 2, and (x + 1/x)² = x² + 1/x² + 2.
• x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x).
• Equivalently x³ − 1/x³ = (x − 1/x)(x² + 1 + 1/x²), which here is 9 × 84.
• x² + 1/x² = 83 gives x − 1/x = 9 and x + 1/x = √85.

Common traps

• Adding 2 instead of subtracting, which leaves √85 and no whole-number route
• Dropping the 3(x − 1/x) term and answering 729
• Ignoring x > 1 and taking the negative root −9

No directly related past PYQ was found.