If $ x = a + m $, $ y = b + m $, $ z = c + m $, the value of $$ \frac{x^2 + y^2 + z^2 - yz - zx - xy}{a^2 + b^2 + c^2 - ab - bc - ca} $$ is

### Concept & Formula This problem centers on a special cyclic algebraic identity. It demonstrates that a specific combination of squares and cross-products is invariant under constant addition. The critical algebraic identity is: $$ x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2}\left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] $$ ### Step-by-Step Solution We are given the equations: 1. $ x = a + m $ 2. $ y = b + m $ 3. $ z = c + m $ Let's calculate the differences between the variables $ x, y, $ and $ z $: $$ x - y = (a + m) - (b + m) = a - b $$ $$ y - z = (b + m) - (c + m) = b - c $$ $$ z - x = (c + m) - (a + m) = c - a $$ Notice that the constant $ m $ cancels out entirely. The difference between the new variables is exactly the same as the difference between the original variables. Now, look at the expression we need to evaluate: $$ \text{Numerator} = x^2 + y^2 + z^2 - xy - yz - zx $$ Apply the cyclic identity to rewrite the numerator: $$ \text{Numerator} = \frac{1}{2}\left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] $$ Substitute the differences we calculated: $$ \text{Numerator} = \frac{1}{2}\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right] $$ Now look at the denominator of the given expression: $$ \text{Denominator} = a^2 + b^2 + c^2 - ab - bc - ca $$ Apply the exact same cyclic identity to rewrite the denominator: $$ \text{Denominator} = \frac{1}{2}\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right] $$ Finally, put the rewritten numerator over the rewritten denominator: $$ \frac{\frac{1}{2}\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]}{\frac{1}{2}\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]} $$ Because the numerator and denominator are identical, the entire fraction simplifies to 1. ### Exam Strategy & Shortcut Use the "Value Putting" method. Since the expression must hold true for *any* values of $ a, b, c, $ and $ m $, pick simple numbers. Let $ a = 1, b = 2, c = 3 $. Let $ m = 1 $. Then $ x = 2, y = 3, z = 4 $. Denominator: $ 1^2 + 2^2 + 3^2 - (2) - (6) - (3) = 14 - 11 = 3 $. Numerator: $ 2^2 + 3^2 + 4^2 - (6) - (12) - (8) = 29 - 26 = 3 $. Result: $ 3 / 3 = 1 $. This practical method bypasses complex algebraic proofs entirely. ### Common Pitfall The biggest mistake is trying to expand the numerator by substituting $ (a + m)^2 $, $ (b + m)^2 $, etc. This creates a massive polynomial with 18+ terms, which is extremely difficult to simplify without making a sign or grouping error. ### Final Answer **Therefore, the correct answer is 1.**