The value of , where x ≠ y ≠ z, is equal to:

Given: 

To find: value of expression $$\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)} , \text{ where} \ x \ne y \ne z .$$ 

Concept Used: 

If a + b + c = 0 or  a = b = c ,

then , $$a^3 +b^3+c^3=3abc$$ 

Solution: 

comparing with  a + b + c = 0,

(x - y) + (y - z) + (z - x) = 0 

 So,

  $$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x) \\ \text{ Now, Putting this in expression}, \\ \implies \frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)} \\ \ \\ \implies \frac{3(x-y)(y-z)(z-x)}{6(x-y)(y-z)(z-x)} \\ \ \\ \implies \frac{\cancel{3(x-y)(y-z)(z-x)}}{2 \times \cancel{3(x-y)(y-z)(z-x)}} \\ \ \\ \implies \frac{1}{2}$$​​

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