The value of $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{9(x-y)(y-z)(z-x)}$ is equal to
(a)0
(b)$\frac{1}{9}$
(c)$\frac{1}{3}$
(d)1
Answer
Answer (as printed): C
Explanation
Since $(x-y)+(y-z)+(z-x)=0$, so $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)$. $$\therefore \quad \text { Given exp. }=\frac{3(x-y)(y-z)(z-x)}{9(x-y)(y-z)(z-x)}=\frac{1}{3} .$$
Explanation as extracted from the printed page; notation may be imperfect.