[ (m - n) 3 + (n - r) 3 + (r - m) 3 ] / 6[(m - n) (n - r) (r - m)] =?

Let, (m - n) = a,
(n -r) = b
(r - m) = c,
Now a + b + c = (m - n) + (n -r) + (r - m)
⇒ a + b + c = m - n + n - r + r - m
⇒ a + b + c = 0............................. (i)
As we know the Algebra formula,
a³ + b³ + c³ ? 3abc = (a+b+c) X 1/2[(a?b)²+(b?c)²+(a?c)²]
Put the value of a + b + c from equation (i).
⇒ a³ + b³ + c³ ? 3abc = 0 X 1/2[(a?b)²+(b?c)²+(a?c)²]
⇒ a³ + b³ + c³ ? 3abc = 0
∴ a³ + b³ + c³ = 3abc
∴ Given expression in question is
[ (m - n)³ + (n - r)³ + (r - m)³ ]/ 6(m - n) (n - r) (r - m)
= ( a³ + b³ + c³ ) / 6abc
= 3abc/6abc
= 1/2