Algebraic identity under the condition a + b + c = 0 Evaluate the product: [(a + b)/c + (b + c)/a + (c + a)/b] × [a/(b + c) + b/(c + a) + c/(a + b)].

Introduction / Context:This algebra question probes symmetry and substitution when a + b + c = 0. Recognizing how this constraint transforms compound fractions allows a short, elegant evaluation without heavy computation.

Given Data / Assumptions:

• a + b + c = 0 (none of a, b, c, or their pairwise sums are zero in denominators unless expressly implied by the identity).
• Target expression: S1 × S2 where S1 = (a + b)/c + (b + c)/a + (c + a)/b and S2 = a/(b + c) + b/(c + a) + c/(a + b).

Concept / Approach:From a + b + c = 0, we have a + b = −c, b + c = −a, and c + a = −b. Substitute these into each fraction to simplify immediately.

Step-by-Step Solution:S1 termwise: (a + b)/c = (−c)/c = −1; (b + c)/a = (−a)/a = −1; (c + a)/b = (−b)/b = −1.Hence S1 = (−1) + (−1) + (−1) = −3.S2 termwise: a/(b + c) = a/(−a) = −1; b/(c + a) = b/(−b) = −1; c/(a + b) = c/(−c) = −1.Hence S2 = (−1) + (−1) + (−1) = −3.Product: S1 × S2 = (−3) * (−3) = 9.

Verification / Alternative check:Pick a simple triple satisfying a + b + c = 0, e.g., a = 1, b = 2, c = −3, and evaluate numerically to confirm S1 = S2 = −3.

Why Other Options Are Wrong:0, 8, −3, and 3 ignore the paired substitutions that force each sum to −3; only the product 9 matches.

Common Pitfalls:Forgetting that b + c = −a, etc.; attempting to find a common denominator instead of using the given linear constraint for direct simplification.

Final Answer:9