Let

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₁→ R₁ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

⇒ Δ = 0 + (2c)[(b)(a + b – c)] + (–2b)[–(c)(c + a – b)]

⇒ Δ = 2bc(a + b – c) + 2bc(c + a – b)

⇒ Δ = 2bc[(a + b – c) + (c + a – b)]

⇒ Δ = 2bc[2a]

∴ Δ = 4abc

Thus,