If $a + b + c = 0$, $(a + b)(b + c)(c + a)$ equals

### Concept & Logic When given a zero-sum condition like $a + b + c = 0$, you can express the sum of any two variables as the negative of the third variable. This simplifies complex products immediately. ### Step-by-Step Solution * **Given:** $a + b + c = 0$. We need to find the value of $(a + b)(b + c)(c + a)$. * **Deduction:** Using the given equation, isolate the pairs: * $a + b = -c$ * $b + c = -a$ * $c + a = -b$ * **Substitution:** Replace the pairs in the target expression with their single-variable equivalents: $$(-c) \times (-a) \times (-b)$$ * **Calculation:** Multiply the terms. The product of three negative numbers remains negative. $$- (c \cdot a \cdot b) = -abc$$ ### Exam Strategy & Shortcut **Value Substitution Method:** You can verify this quickly by plugging in simple numbers that satisfy $a + b + c = 0$. Let $a = 1$, $b = 1$, $c = -2$. Expression: $(1 + 1)(1 - 2)(-2 + 1) = (2)(-1)(-1) = 2$. Now test the options: (a) $ab(a+b) = (1)(1)(2) = 2$ (Wait, let's check if this holds for other values). (c) $-abc = -(1)(1)(-2) = 2$. If multiple options yield the same result, pick a different set of numbers (e.g., $1, 2, -3$) to break the tie. However, the direct algebraic substitution shown in the step-by-step solution is actually faster and foolproof here. ### Common Pitfall Multiplying out the brackets $(a+b)(b+c)(c+a)$ to get $a^2b + ab^2 + ...$ is a massive time sink. Always look for ways to use the constraint equation ($a+b+c=0$) to reduce terms before multiplying. ### Final Answer Therefore, the correct answer is **-abc**.