If x + 1⁄x = 2√10, where x > 1, then the value of (x³ − 1⁄x³) is:

Correct — B. The stem is printed as an image and reads: if x + 1⁄x = 2√10, where x > 1, then the value of (x³ − 1⁄x³) is.

Rule: (x − 1⁄x)² = (x + 1⁄x)² − 4
= (2√10)² − 4 = 40 − 4 = 36
x − 1⁄x = 6, taking the positive root because x > 1

Then x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x)
= 6³ + 3 × 6 = 216 + 18 = 234 → option (b)

Why the others are wrong

• (a)
  198 — 198 is 216 − 18, the middle term subtracted instead of added. For a difference of cubes the correction 3(x − 1⁄x) carries a plus sign.
• (c)
  216 — 216 is the bare cube 6³. It stops at (x − 1⁄x)³ and drops the +3(x − 1⁄x) term, which is worth 18 here.
• (d)
  221 — 221 matches no step of the working. With x − 1⁄x = 6 the identity can only deliver 216 before the correction and 234 after it.

Two identities do all the work in the x + 1⁄x family, and both come from the product x × 1⁄x = 1.

Squaring: (x + 1⁄x)² and (x − 1⁄x)² differ by exactly 4, because the cross term is 2 either way.

Cubing: a³ − b³ = (a − b)³ + 3ab(a − b), and with ab = 1 that becomes x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x).

So a sum is converted into a difference, and the difference is cubed. The range given for x is not decoration — it chooses the sign of the square root.

Key facts

• (x + 1⁄x)² − (x − 1⁄x)² = 4 for every non-zero x.
• x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x).
• x > 1 forces x − 1⁄x to be positive, which is why 6 is taken rather than −6.
• Here (2√10)² = 40, so x − 1⁄x = 6 and x³ − 1⁄x³ = 234.

Common traps

• Cubing 2√10 straight away and trying to reach a difference of cubes from a sum
• Answering −234 by taking the negative root when the stem has already fixed x > 1

No directly related past PYQ was found.