For real numbers a, b, and c, the expression a^2 + b^2 + c^2 - ab - bc - ca equals zero if and only if which of the following conditions holds?

Introduction / Context:
This algebra question tests understanding of symmetric expressions in three variables and the use of identities to factor or simplify them. The expression a^2 + b^2 + c^2 - ab - bc - ca may look complicated, but with some algebraic manipulation it can be rewritten in a way that reveals when it becomes zero. Recognizing such patterns is very helpful in inequality problems and in simplifying expressions in algebra and coordinate geometry.

Given Data / Assumptions:
- We consider real numbers a, b, and c.
- The expression in question is E = a^2 + b^2 + c^2 - ab - bc - ca.
- We want to know under what condition E = 0 holds.
- Options involving products like abc = 1 or abc = 0 are also given, but we must check them carefully rather than assume multiplicative relationships.

Concept / Approach:
There is a well-known identity that rewrites a^2 + b^2 + c^2 - ab - bc - ca as a sum of squares involving pairwise differences. Specifically, E can be expressed as ((a - b)^2 + (b - c)^2 + (c - a)^2) / 2. Since squares of real numbers are always non-negative, E is always greater than or equal to zero. The only way E can be zero is if each squared term is zero, i.e., all pairwise differences vanish and a, b, c are all equal. This directly leads to the required condition.

Step-by-Step Solution:
Step 1: Start from the identity for pairwise differences:
(a - b)^2 = a^2 + b^2 - 2ab,
(b - c)^2 = b^2 + c^2 - 2bc,
(c - a)^2 = c^2 + a^2 - 2ca. Step 2: Add these three expressions:
(a - b)^2 + (b - c)^2 + (c - a)^2 = (a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (c^2 + a^2 - 2ca). Step 3: Combine like terms: we get 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca. Step 4: Factor out 2: this becomes 2(a^2 + b^2 + c^2 - ab - bc - ca). Step 5: Therefore a^2 + b^2 + c^2 - ab - bc - ca = ((a - b)^2 + (b - c)^2 + (c - a)^2) / 2. Step 6: A sum of squares equals zero if and only if each square is zero. So (a - b)^2 = 0, (b - c)^2 = 0, and (c - a)^2 = 0, meaning a = b, b = c, and c = a. Step 7: These equalities together imply a = b = c.

Verification / Alternative check:
If a = b = c = k, then E = k^2 + k^2 + k^2 - k^2 - k^2 - k^2 = 0, confirming that equality of the three numbers makes the expression zero. On the other hand, if a, b, c are not all equal, at least one difference such as a - b or b - c is non-zero, making at least one squared term positive. Then the sum of squares is positive and E is greater than zero. Thus E = 0 occurs if and only if a = b = c.

Why Other Options Are Wrong:
Option abc = 1 or abc = -1: These conditions relate to the product of a, b, and c and say nothing about pairwise equality. There are many triples with product 1 or -1 for which a^2 + b^2 + c^2 - ab - bc - ca is not zero.
Option abc = 0: This means at least one of the numbers is zero, but the others need not match. E is generally positive for such choices unless they all coincide, which is a special case of a = b = c.
Option None of these: Incorrect because the condition a = b = c is both necessary and sufficient for the expression to be zero and is listed explicitly.

Common Pitfalls:
Students often try to factor the expression incorrectly or look for multiplicative conditions like abc = 1 without checking them. Others might test only a few numerical examples and jump to conclusions. Using the identity that expresses the expression as a sum of squared differences is the most reliable way to understand its behaviour and to see that equality of all three variables is the exact condition for it to vanish.

Final Answer:
The expression equals zero if and only if a, b, and c are all equal, that is, a = b = c.