The value of $$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}$$ is

### Concept & Formula This fraction simplifies using the conditional identity for the sum of cubes: If $A + B + C = 0$, then $A^3 + B^3 + C^3 = 3ABC$. ### Step-by-Step Solution * Let the numerator components be $A = x - y$, $B = y - z$, and $C = z - x$. * Notice that $A + B + C = (x - y) + (y - z) + (z - x) = 0$. * Therefore, the numerator becomes $3(x - y)(y - z)(z - x)$. * Now, let the denominator components be $P = x^2 - y^2$, $Q = y^2 - z^2$, and $R = z^2 - x^2$. * Notice that $P + Q + R = (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0$. * Therefore, the denominator becomes $3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)$. * Expand the denominator using the difference of squares $a^2 - b^2 = (a-b)(a+b)$: $$3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)$$ * Divide the simplified numerator by the simplified denominator: $$\frac{3(x - y)(y - z)(z - x)}{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}$$ * Cancel the common terms $3(x - y)(y - z)(z - x)$ from top and bottom: $$= \frac{1}{(x + y)(y + z)(z + x)}$$ * Rewrite using a negative exponent: $$= [(x + y)(y + z)(z + x)]^{-1}$$ ### Exam Strategy & Shortcut Whenever you see a sum of three cubes in a pattern that looks cyclical (like $x-y, y-z, z-x$), immediately check if they sum to zero. If they do, instantly replace them with 3 times their product. This visual recognition avoids lengthy expansions. ### Common Pitfall A common error is to try expanding $(x - y)^3$ individually. This creates a massive, unmanageable polynomial that guarantees a loss of time and inevitable calculation errors. ### Final Answer Therefore, the correct answer is **$[(x + y) (y + z) (z + x)]^{-1}$**.