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 :
Answer & explanation
Correct answer: option 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
$\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$ = $\frac{(2-1)^3+(1-0)^3+(0-2)^3}{6(2-1)(1-0)(0-2)}$
= \(\frac{1 + 1 - 8}{6 × -2}\) = \(\frac{-6}{-12}\)
= $\frac{1}{2}$