Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{9}$
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Correct Answer: $\frac{1}{2}$
Solution : Given expression, $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$ Let $x-y = a,$ $y-z = b,$ and $z-x = c$ ⇒ The given expression becomes, $\frac{a^3+b^3+c^3}{6abc}$ Now, $a+b+c= x-y+y-z+z-x = 0$ ⇒ When $a+b+c=0,\ a^3 + b^3 + c^3 = 3abc$ ⇒ $\frac{a^3+b^3+c^3}{6abc} = \frac{3abc}{6abc}=\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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