L.H.S =

Apply C₁→C₁ + C₂ + C₃

Taking (a + b + c) common from C₁ we get,

Applying, R₃→R₃ – 2R₁

= (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)]

= a³ + b³ + c³ – 3abc

As, L.H.S = R.H.S

Hence, proved.