If 1/a + 1/b + 1/c = 1/(a + b + c) and a + b + c ≠ 0 and abc ≠ 0, what is the value of (a + b)(b + c)(a + c)?

Introduction / Context:
This algebraic identity question involves reciprocals and symmetric expressions in three variables a, b, and c. We are given a special condition relating 1/a, 1/b, and 1/c to 1/(a + b + c) and asked to find the value of the product (a + b)(b + c)(a + c). Problems like this strengthen understanding of algebraic manipulation and standard identities.

Given Data / Assumptions:

• 1/a + 1/b + 1/c = 1/(a + b + c).
• a + b + c ≠ 0.
• abc ≠ 0, so a, b, and c are all nonzero.
• We need to determine (a + b)(b + c)(a + c).

Concept / Approach:
First rewrite the given reciprocal condition with a common denominator to express it in terms of ab + bc + ca and abc. Then use a known identity that connects (a + b)(b + c)(c + a) with (a + b + c)(ab + bc + ca) and abc. Specifically, there is an identity: (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc. Once we find ab + bc + ca in terms of abc and a + b + c using the given condition, we can substitute into this identity to find the required value.

Step-by-Step Solution:
Step 1: Start from the given condition: 1/a + 1/b + 1/c = 1/(a + b + c).Step 2: Write the left side with a common denominator: 1/a + 1/b + 1/c = (bc + ac + ab) / (abc).Step 3: So (bc + ac + ab) / (abc) = 1/(a + b + c).Step 4: Cross multiply: (bc + ac + ab)(a + b + c) = abc.Step 5: Recognise that bc + ac + ab is the symmetric sum ab + bc + ca.Step 6: Therefore (ab + bc + ca)(a + b + c) = abc.Step 7: Recall the identity: (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc.Step 8: From Step 6, we have (a + b + c)(ab + bc + ca) = abc.Step 9: Substitute this into the identity: (a + b)(b + c)(c + a) = abc - abc = 0.Step 10: Therefore (a + b)(b + c)(a + c) = 0.

Verification / Alternative check:
To verify concretely, we can choose specific values of a, b, and c that satisfy the given condition. For instance, try a = 1, b = 1, and c = -1. Then a + b + c = 1, which is nonzero. Now compute 1/a + 1/b + 1/c = 1/1 + 1/1 + 1/(-1) = 1 + 1 - 1 = 1. Also, 1/(a + b + c) = 1/1 = 1. So the condition holds. Now compute (a + b)(b + c)(a + c) = (1 + 1)(1 - 1)(1 - 1) = 2 * 0 * 0 = 0. This matches our algebraic result, confirming the reasoning.

Why Other Options Are Wrong:
Option 3, 2, 1, and -1: These suggest that the product (a + b)(b + c)(a + c) might be some nonzero constant, but our algebra clearly shows it reduces to abc - abc = 0 for any a, b, and c satisfying the given condition.

Common Pitfalls:
Students may incorrectly expand (a + b)(b + c)(c + a) or forget the standard identity connecting it with (a + b + c)(ab + bc + ca). Others might mishandle the reciprocals, especially when finding the common denominator. Careful algebraic manipulation and remembering common symmetric identities help avoid these errors. When in doubt, testing specific numeric examples consistent with the condition is a good way to build confidence in the result.

Final Answer:
The value of (a + b)(b + c)(a + c) is 0.