The value of $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{\left(x^{2}-y^{2}\right)^{3}+\left(y^{2}-z^{2}\right)^{3}+\left(z^{2}-x^{2}\right)^{3}}$ is
(a)0
(b)1
(c)$[2(x+y+z)]^{-}$
(d)$[(x+y)(y+z)(z+x)]^{-}$
Answer
Answer (as printed): D
Explanation
Since $(x-y)+(y-z)+(z-x)=0$, so $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)$. Since $\left(x^{2}-y^{2}\right)+\left(y^{2}-z^{2}\right)+\left(z^{2}-x^{2}\right)=0$, so $\left(x^{2}-y^{2}\right)^{3}+\left(y^{2}-z^{2}\right)^{3}+\left(z^{2}-x^{2}\right)^{3}=3\left(x^{2}-y^{2}\right)\left(y^{2}-z^{2}\right)$ $\left(z^{2}-x^{2}\right)$. $$\begin{aligned} \therefore \text { Given exp. } & =\frac{3(x-y)(y-z)(z-x)}{3\left(x^{2}-y^{2}\right)\left(y^{2}-z^{2}\right)\left(z^{2}-x^{2}\right)} \\ & =\frac{1}{(x+y)(y+z)(z+x)} \\ & =[(x+y)(y+z)(z+x)]^{-} \end{aligned}$$