In the given expression, we see that both numerator and denominator is in the form a3 + b3 + c3 = 3abc because a + b + c = 0.
From numerator, we see that
a2 – b2 + b2 – c2 + c2 – a2 = 0
⇒ (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3
= 3 (a2 – b2)(b2 – c2)(c2 – a2)
Similarly, from denominator,
a – b + b – c + c – a = 0
(a – b)3 + (b – c)3 + (c – a)3
= 3(a – b)(b – c)(c – a)
Value of the expression
