Evaluate the following:

Applying, C₁→C₁ + C₂ + C₃, we get,

Taking, (a + b + c) we get,

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

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

= (a + b + c)[a² – ac – ab + bc + b² + c² – 2bc]

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

So, Δ = (a + b + c)[a² + b² + c² – ac – ab – bc]