Applying, C₂→C₂ – 2C₁ – 2C₃, we get,

Taking, – (a² + b² + c²) common from C₂ we get,

Applying R₂→R₂ – R₁ and R₃→R₃ – R₁, we get

= – (a² + b² + c²)(a – b)(c – a)[( – (b + a))( – b) – (c)(c + a)]

= (a – b)(b – c)(c – a)(a + b + c)(a² + b² + c²)

= R.H.S

Hence, proved.