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

Taking (a + b) and (c + a) common from C₂ and C₃, we get

Applying R₂→ R₂ + R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding the determinant along C₃, we have

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

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

⇒ Δ = (a + b)(c + a)(2b + 2c)

∴ Δ = 2(a + b)(b + c)(c + a)

Thus,