Standard cube-sum identity with cyclic differences (minimal repair): Let x = a − b, y = b − c, z = c − a. Evaluate x^3 + y^3 + z^3 − 3xyz.

Introduction / Context:This classic identity says x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). When x + y + z = 0, the entire expression equals 0. With cyclic differences x = a − b, y = b − c, z = c − a, the sum vanishes automatically.

Given Data / Assumptions:

• x = a − b, y = b − c, z = c − a.
• a, b, c are real numbers.

Concept / Approach:Compute x + y + z: it telescopes to 0. Then apply the identity to conclude the target expression is 0, without needing the individual values of a, b, c.

Step-by-Step Solution:

Verification / Alternative check:Choose specific values, e.g., a = 3, b = 2, c = 1 ⇒ x = 1, y = 1, z = −2. Compute 1 + 1 − 8 − 3*(1*1*−2) = −6 + 6 = 0.

Why Other Options Are Wrong:a + b + c, 3abc, or scaled sums require nonzero (x + y + z); here it is 0, forcing the whole expression to 0.

Common Pitfalls:Forgetting to test x + y + z first; the identity simplifies dramatically once you notice the telescoping sum.

Final Answer:0