Key identity for sums of cubes: Given real a, b, c with a^3 + b^3 + c^3 = 3abc and a + b + c ≠ 0, determine the relation among a, b, c.

Introduction / Context:
The identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca) characterizes when the sum of cubes equals thrice the product. This problem tests recognition of the special cases implied by this factorization.

Given Data / Assumptions:

• a^3 + b^3 + c^3 = 3abc.
• a + b + c ≠ 0.
• a, b, c are real numbers.

Concept / Approach:
From the identity, if a^3 + b^3 + c^3 − 3abc = 0, then either a + b + c = 0, or a^2 + b^2 + c^2 − ab − bc − ca = 0. The second factor equal to zero implies a = b = c for real numbers because it can be written as 1/2[(a − b)^2 + (b − c)^2 + (c − a)^2].

Step-by-Step Solution:

Verification / Alternative check:
If a = b = c, then a^3 + b^3 + c^3 = 3a^3 and 3abc = 3a^3, so the equality holds for any real a (and the sum a + b + c = 3a is nonzero if a ≠ 0).

Why Other Options Are Wrong:
Linear relations like a + b = c, a + c = b, or b + c = a do not, in general, force the factor a^2 + b^2 + c^2 − ab − bc − ca to vanish.

Common Pitfalls:
Forgetting the factorization or assuming a + b + c = 0 must hold; here it is explicitly ruled out, forcing a = b = c instead.

Final Answer:
a = b = c