Prove the following identities –

= (a + b - c)(b + c - a)(c + a - b)

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

Taking the term (a – b – c) common from R₁, we get

Applying C₂→ C₂ + C₁, we get

Applying C₃→ C₃ + C₁, we get

Expanding the determinant along R₁, we have

Δ = (a – b – c)[–1(b + a – c)(c + a – b) – 0 + 0]

⇒ Δ = –(a – b – c)(b + a – c)(c + a – b)

∴ Δ = (b + c – a)(a + b – c)(c + a – b)

Thus,