Using properties of determinants, prove the following:

Given,

Performing the operation C₁→ C₁ – C₂ and C₂→ C₂ – C₃, we get

We know that a³ – b³ = (a – b) (a² + ab + b²)

Taking (a – b) and (b – c) common from C₁ and C₂ respectively, we get

Expanding along R₁, we get

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

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

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

= (a – b) (b – c) (c – a) (c + a + b)

= (a – b) (b – c) (c – a) (a + b + c)

= RHS

Hence proved.