Prove the followi

Let

Taking a, b and c common from C₁, C₂ and C₃, 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 R₁→ R₁ – R₂, we get

Taking the term (a² + b² + c²) common from R₁, we get

Applying R₂→ R₂ – R₃, we get

Taking the term (a² + b² + c²) common from R₂, we get

Applying C₂→ C₂ + C₁, we get

Expanding the determinant along R₁, we have

Δ = (abc)(a² + b² + c²)²(–1)[(a² + b² – c²) – (2b² + 2a²)]

⇒ Δ = (abc)(a² + b² + c²)²[–(a² + b² – c²) + (2b² + 2a²)]

⇒ Δ = (abc)(a² + b² + c²)²[–a² – b² + c² + 2b² + 2a²]

⇒ Δ = (abc)(a² + b² + c²)²[a² + b² + c²]

∴ Δ = (abc)(a² + b² + c²)³

Thus,