Prove the following identities –

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 C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = (x + y + z)(1)[0 – (–(x + y + z)(x + y + z))]

⇒ Δ = (x + y + z)(x + y + z)(x + y + z)

∴ Δ = (x + y + z)³

Thus,