Evaluate the expression:(1⁄(3 − √8)) − (1⁄(√8 − √7)) + (1⁄(√7 − √6)) − (1⁄(√6 − √5)) + (1⁄(√5 − 2)) = ?

Correct — A. Rationalise each term with its conjugate. Write 3 as √9 and 2 as √4: in every denominator the two numbers under the roots then differ by 1, so each denominator becomes 1.

1⁄(3 − √8) = (3 + √8)⁄(9 − 8) = 3 + √8
1⁄(√8 − √7) = √8 + √7
1⁄(√7 − √6) = √7 + √6

1⁄(√6 − √5) = √6 + √5
1⁄(√5 − 2) = (√5 + 2)⁄(5 − 4) = √5 + 2

Put the signs back:
(3 + √8) − (√8 + √7) + (√7 + √6) − (√6 + √5) + (√5 + 2)
Each root appears once added and once subtracted, so all cancel.
Left: 3 + 2 = 5 → option (a)

Why the others are wrong

• (b)
  3 — 3 is the first term's whole part alone. The last term, 1⁄(√5 − 2) = √5 + 2, adds a 2 that nothing cancels.
• (c)
  2 — 2 is what the last term leaves on its own. The first term, 1⁄(3 − √8) = 3 + √8, adds a 3 that no later term removes.
• (d)
  0 — It assumes everything cancels. The roots do, but the 3 at the start and the 2 at the end have no partner of opposite sign.

The conjugate of √a − √b is √a + √b. Their product is a − b, with no root left, because (x − y)(x + y) = x² − y².

So 1⁄(√a − √b) = (√a + √b)⁄(a − b), and when a − b = 1 the reciprocal is simply √a + √b.

The sum then telescopes: with alternating signs each middle root is added once and subtracted once, and only the two ends survive.

Key facts

• 1⁄(√(n + 1) − √n) = √(n + 1) + √n, because (√(n + 1) − √n)(√(n + 1) + √n) = 1.
• 3 = √9 and 2 = √4, so 3 − √8 and √5 − 2 follow the same pattern as the middle terms.
• (x − y)(x + y) = x² − y² is the identity behind every conjugate step.

Common traps

• Keeping the minus sign: 1⁄(3 − √8) is 3 + √8, not 3 − √8. Rationalising flips the sign between the two parts.
• Not recognising 3 and 2 as √9 and √4, and so missing that the first and last terms fit the same pattern.

No directly related past PYQ was found.