If a³ + 1⁄a³ = 2 (a > 0), then the value of a + 1⁄a is:

Correct — A. Put t = a + 1⁄a and use the cube identity.

(a + 1⁄a)³ = a³ + 1⁄a³ + 3(a + 1⁄a)
t³ = 2 + 3t
t³ − 3t − 2 = 0

Factorise: (t − 2)(t + 1)² = 0, so t = 2 or t = −1.
a > 0 forces a + 1⁄a ≥ 2, which kills t = −1.
t = 2 → option (a).

Shortest route: for a > 0 the sum a³ + 1⁄a³ is smallest at a = 1, where it is exactly 2. So a = 1 and a + 1⁄a = 2.

Why the others are wrong

• (b)
  4 — Test 4 against the given equation: if a + 1⁄a were 4, then a³ + 1⁄a³ = 4³ − 3 × 4 = 52, not 2.
• (c)
  1 — 1 is out of range before any algebra. For every a > 0, AM–GM gives a + 1⁄a ≥ 2, with equality only at a = 1.
• (d)
  3 — 3 gives a³ + 1⁄a³ = 3³ − 3 × 3 = 18, not 2. Only t = 2 satisfies t³ − 3t = 2 while keeping a positive.

Two identities carry every question of this shape.

(a + 1⁄a)³ = a³ + 1⁄a³ + 3(a + 1⁄a) — knowing the cube sum turns the question into a cubic in t = a + 1⁄a.

a² + 1⁄a² = t² − 2 is the same trick one power down, and it is what SSC asks when the given value is a square sum.

The condition a > 0 is not decoration. The cubic offers t = −1 as a double root, and only positivity removes it: a + 1⁄a is at least 2 for every positive a and at most −2 for every negative one.

Key facts

• (a + 1⁄a)³ = a³ + 1⁄a³ + 3(a + 1⁄a).
• For a > 0, a + 1⁄a ≥ 2, with equality exactly at a = 1.
• t³ − 3t − 2 factorises as (t − 2)(t + 1)², so its roots are 2 and −1.
• a³ + 1⁄a³ = 2 with a > 0 has the single solution a = 1.

Common traps

• Dropping the 3(a + 1⁄a) term from the cube expansion and solving t³ = 2 + 3
• Taking t = −1 from the cubic because the algebra allows it, and ignoring a > 0
• Cubing the answer to check and comparing it with 2 instead of using t³ − 3t

No directly related past PYQ was found.