The value of $$ \frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{9 (x - y) (y - z) (z - x)} $$ is equal to

### Concept & Formula This problem leverages a highly specific conditional case of the three-variable sum of cubes formula. The core conditional identity is: If $ A + B + C = 0 $, then it is always true that: $$ A^3 + B^3 + C^3 = 3ABC $$ ### Step-by-Step Solution Let's define new variables to represent the binomials in the expression: * Let $ A = (x - y) $ * Let $ B = (y - z) $ * Let $ C = (z - x) $ First, let's check the condition $ A + B + C $ by adding these terms together: $$ A + B + C = (x - y) + (y - z) + (z - x) $$ $$ = x - y + y - z + z - x $$ Notice that all the variables perfectly cancel each other out: $$ = 0 $$ Since $ A + B + C = 0 $ is definitively true, we can apply the conditional identity: $$ A^3 + B^3 + C^3 = 3ABC $$ Substitute the original binomials back into this identity: $$ (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) $$ Now, substitute this simplified result back into the numerator of the original problem: $$ \text{Expression} = \frac{3(x - y)(y - z)(z - x)}{9(x - y)(y - z)(z - x)} $$ The entire block $ (x - y)(y - z)(z - x) $ is identical in both the top and bottom. Cancel it out completely: $$ = \frac{3}{9} $$ Simplify the remaining fraction: $$ = \frac{1}{3} $$ ### Exam Strategy & Shortcut Recognize the "cyclic format". Whenever you see expressions looping sequentially like $ (x-y) $, $ (y-z) $, $ (z-x) $, their sum is zero. Therefore, their cubes add up to three times their product. Seeing $ \frac{3 \times \text{Product}}{9 \times \text{Product}} $ allows you to instantly bypass all writing and evaluate $ 3/9 = 1/3 $ in a matter of seconds. ### Common Pitfall A major mistake is attempting to individually expand $ (x - y)^3 $, $ (y - z)^3 $, and $ (z - x)^3 $. This generates 12 separate algebraic terms, creating an overwhelming polynomial that is highly susceptible to sign errors and takes entirely too long to manually cancel out. Always look for zero-sum conditions first. ### Final Answer **Therefore, the correct answer is $ \frac{1}{3} $.**