Given 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:Algebraic manipulations often simplify dramatically when a + b + c = 0. This question checks whether you can exploit that condition to reduce complicated rational expressions. Recognizing that sums like a + b equal −c is the main lever to collapse each fraction to simple constants.

Given Data / Assumptions:

• a + b + c = 0 with non-zero a, b, c to keep denominators valid.
• Compute the product of two bracketed sums.

Concept / Approach:Use the identity implied by the condition: a + b = −c, b + c = −a, c + a = −b. Substituting these into all numerators and denominators instantly converts each fraction into −1, which makes each bracket sum straightforward. Finally, multiply the two summed values.

Step-by-Step Solution:Since a + b = −c, (a + b)/c = (−c)/c = −1.Similarly, (b + c)/a = (−a)/a = −1, and (c + a)/b = (−b)/b = −1.Therefore the first bracket equals (−1) + (−1) + (−1) = −3.In the second bracket: a/(b + c) = a/(−a) = −1; b/(c + a) = b/(−b) = −1; c/(a + b) = c/(−c) = −1.So the second bracket also equals −3.Product = (−3) * (−3) = 9.

Verification / Alternative check:Pick a concrete triple with sum zero, e.g., a = 1, b = 2, c = −3. Substitute to verify each fraction is −1. The product again equals 9.

Why Other Options Are Wrong:

• 0, 8, −3, 3: These values come from arithmetic slips or forgetting that both brackets sum to −3.

Common Pitfalls:Attempting to find a common denominator instead of exploiting a + b + c = 0; sign errors when converting a + b to −c.

Final Answer:9