Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}$, where $x \neq y \neq z$, is:
Option 1: $0$
Option 2: $\frac{1}{(x+y+z)}$
Option 3: $\frac{1}{(x+y)(y+z)(z+x)}$
Option 4: $1$
Correct Answer: $\frac{1}{(x+y)(y+z)(z+x)}$
Solution :
Given:
$\frac{(x-y)^3+(y-z)^3+(z-x)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}$
= $\frac{3(x-y)(y-z)(z-x)}{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}$ [since $(x-y)+(y-z)+(z-x)=0$]
= $\frac{(x-y)(y-z)(z-x)}{(x+y)(x-y)(y+z)(y-z)(z+x)(z-x)}$
= $\frac{1}{(x+y)(y+z)(z+x)}$
Hence, the correct answer is $\frac{1}{(x+y)(y+z)(z+x)}$.
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