Cube–reciprocal identity with a = (√2 − 1)^(1/3) Evaluate (a − a^−1)^3 + 3(a − a^−1).

Introduction / Context:
This problem tests recognition of a compact algebraic identity that relates (a − 1/a)^3 + 3(a − 1/a) to a^3 − 1/a^3. Using the given a = (√2 − 1)^(1/3), we can evaluate the expression exactly without messy decimals.

Given Data / Assumptions:

• a = (√2 − 1)^(1/3).
• Target: (a − a^−1)^3 + 3(a − a^−1).
• All algebra performed with exact radicals.

Concept / Approach:
Identity: (t)^3 + 3t = a^3 − 1/a^3 where t = a − 1/a. Expand (a − 1/a)^3 to verify this, or use the well-known result to avoid expansion. Then compute a^3 and a^−3 directly from the definition of a.

Step-by-Step Solution:
Let t = a − 1/a. Then t^3 + 3t = a^3 − 1/a^3.Given a = (√2 − 1)^(1/3) ⇒ a^3 = √2 − 1.Compute a^−3 = 1/(√2 − 1) = (√2 + 1)/[(√2 − 1)(√2 + 1)] = √2 + 1.Hence a^3 − a^−3 = (√2 − 1) − (√2 + 1) = −2.

Verification / Alternative check:
Expand (a − 1/a)^3 = a^3 − 3a + 3/a − 1/a^3. Adding 3(a − 1/a) gives a^3 − 1/a^3 indeed, confirming the identity.

Why Other Options Are Wrong:
2, 2√2, and √2 result from sign or reciprocal mistakes; 0 would require a^3 = 1/a^3, which is false here.

Common Pitfalls:
Forgetting to rationalize 1/(√2 − 1), or misapplying the identity by using plus instead of minus in the reciprocal term.

Final Answer:
-2