If a^3 + 1/a^3 = 2 (with a ≠ 0), find the exact value of: a + 1/a Use an algebraic identity to relate a^3 + 1/a^3 to a + 1/a.

Introduction / Context:
This question tests a common identity that connects a + 1/a with a^3 + 1/a^3. Instead of solving for a directly (which could be messy), you use the identity: (a + 1/a)^3 = a^3 + 1/a^3 + 3(a + 1/a). This identity is derived from expanding (u + v)^3 with u = a and v = 1/a, noting that uv = 1. The given value a^3 + 1/a^3 = 2 allows you to create a cubic equation in t = a + 1/a. Then you solve that cubic, which in this case has an easy integer root. These problems are designed so that a simple root like 2 or -2 works, and the correct option appears in the list. The key is to set up the equation correctly and avoid sign or coefficient mistakes, especially the “+ 3t” term that comes from 3uv(u+v) with uv = 1.

Given Data / Assumptions:

• a^3 + 1/a^3 = 2 • a ≠ 0 • Let t = a + 1/a • Identity: (a + 1/a)^3 = a^3 + 1/a^3 + 3(a + 1/a)

Concept / Approach:
Substitute t = a + 1/a into the identity: t^3 = (a^3 + 1/a^3) + 3t. Given a^3 + 1/a^3 = 2, we get t^3 = 2 + 3t, or t^3 − 3t − 2 = 0. Solve this cubic by testing the small integer options (since the question is multiple-choice and usually designed for a clean root).

Step-by-Step Solution:
1) Let t = a + 1/a. 2) Use the identity: t^3 = a^3 + 1/a^3 + 3t 3) Substitute the given value a^3 + 1/a^3 = 2: t^3 = 2 + 3t 4) Rearrange to form a cubic equation: t^3 − 3t − 2 = 0 5) Test t = 2: 2^3 − 3*2 − 2 = 8 − 6 − 2 = 0 6) Since t = 2 satisfies the equation: a + 1/a = 2

Verification / Alternative check:
If a + 1/a = 2, then multiplying by a gives a^2 − 2a + 1 = 0, so (a − 1)^2 = 0 and a = 1. Then a^3 + 1/a^3 = 1 + 1 = 2, matching the given condition exactly. This confirms that t = 2 is not only an algebraic root but also consistent with a valid nonzero value of a.

Why Other Options Are Wrong:
• 1, 3, 4: do not satisfy t^3 − 3t − 2 = 0 when substituted. • -2: gives (-8) + 6 − 2 = -4, not zero, so it is not a solution here.

Common Pitfalls:
• Forgetting the +3t term in the identity. • Setting t = a − 1/a by mistake. • Arithmetic errors when testing candidate roots.

Final Answer:
2