Prove the following identities –

Let

Taking a², b² and c² common from C₁, C₂ and C₃, we get

Taking a, b and c common from R₁, R₂ and R₃, we get

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 C₂→ C₂ – C₃, we get

Expanding the determinant along R₁, we have

Δ = (a³b³c³)[0 – 0 + 1(1)(1) – (1)(–1)]

⇒ Δ = (a³b³c³)[1 + 1]

∴ Δ = 2a³b³c³

Thus,