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 (x + y + z) common from R₁, we get

Applying C₁→ C₁ – C₂, we get

Applying C₁→ C₁ – C₃, we get

Expanding the determinant along C₁, we have

Δ = (x + y + z)(x – z)[(1)(x) – (z)(1)]

⇒ Δ = (x + y + z)(x – z)(x – z)

∴ Δ = (x + y + z)(x – z)²

Thus,