Difficulty: Easy
Correct Answer: a = b = c
Explanation:
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:
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:
Given a^3 + b^3 + c^3 − 3abc = 0Factor: (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca) = 0Since a + b + c ≠ 0, we must have a^2 + b^2 + c^2 − ab − bc − ca = 0Rewrite: a^2 + b^2 + c^2 − ab − bc − ca = 1/2[(a − b)^2 + (b − c)^2 + (c − a)^2]This sum of squares equals 0 ⇒ a − b = b − c = c − a = 0 ⇒ a = b = cVerification / 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
Discussion & Comments