Simplify the following equation.
​$$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{9(x - y)(y - z)(z - x)} = ?$$​​

Correct option is A

Given:

​$$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{9(x - y)(y - z)(z - x)}$$​

Formula Used:

This equation involves a sum of cubes and can be simplified using the identity:

​$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$​

If a + b + c = 0, the equation simplifies to:

​$$a^3 + b^3 + c^3 = 3abc$$​

Solution:

In this case, take:

a=x−y

b=y−z

c=z−x

Now, observe that:

(x - y) + (y - z) + (z - x) = 0

Since the sum is zero, we can apply the identity for the sum of cubes:

​$$(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$$

Now,

​$$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{9(x - y)(y - z)(z - x)}=\frac{3(x - y)(y - z)(z - x)}{9(x - y)(y - z)(z - x)}$$​

​$$=\frac{3}{9} = \frac{1}{3}$$​