The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where x ≠ y ≠ z, is equal to :

$\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{9}$

$\frac{1}{2}$

$\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$  Put x = 2  y = 1 and  z = 0 $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$ = $\frac{(2-1)^3+(1-0)^3+(0-2)^3}{6(2-1)(1-0)(0-2)}$  = \(\frac{1 + 1 - 8}{6 × -2}\) = \(\frac{-6}{-12}\) = $\frac{1}{2}$