Find the value of $$\frac{(p-q)^3+(q-r)^3+(r-p)^3}{9(p-q)(q-r)(r-p)}$$​.

Correct option is D

Given:

​$$\frac{(p - q)^3 + (q - r)^3 + (r - p)^3}{9(p - q)(q - r)(r - p)}$$​​

Concept Used:

​$$a^3 + b^3 + c^3 = 3abc \quad \text{when} \quad a + b + c = 0$$​​

​$$\frac{a^3 + b^3 + c^3}{9abc} = \frac{3abc}{9abc} = \frac{1}{3}$$​​

Solution:

Let:

​$$a = (p - q), \quad b = (q - r), \quad c = (r - p)$$​​

Then:

a + b + c = (p - q) + (q - r) + (r - p) = 0

​$$\frac{(p - q)^3 + (q - r)^3 + (r - p)^3}{9(p - q)(q - r)(r - p)} = \frac{1}{3}$$​