Algebra under the condition a + b + c = 0: Find the multiplier k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).

Introduction / Context:
This question checks a well-known identity that holds when the linear sum a + b + c equals zero. It ties the sum of squares to the sum of pairwise products, a relationship that often appears in symmetric algebra and factorization problems.

Given Data / Assumptions:

• a + b + c = 0
• We seek k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).

Concept / Approach:
Start with (a + b + c)^2. Expanding yields a^2 + b^2 + c^2 + 2(ab + bc + ca). Under the given condition, the left side is 0, which directly relates the two target sums and allows k to be read off immediately after rearrangement.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 2, 1, 0: None satisfy the derived identity for generic nonzero triples with a + b + c = 0.

Common Pitfalls:

• Losing the sign on the 2(ab + bc + ca) term during rearrangement.
• Assuming k must be positive; here it is negative.

Final Answer: