Prove the followi

Let

Multiplying a, b and c to R₁, R₂ and R₃, we get

Dividing C₁, C₂ and C₃ with a, b and c, 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

Applying R₁→ R₁ – R₃, we get

Applying C₂→ C₂ – C₁, we get

Applying C₃→ C₃ – C₁, we get

Expanding the determinant along R₁, we have

Δ = 0 + (2c²)[(b²)(a² + b² – c²)] + (–2b²)[–(c²)(c² + a² – b²)]

⇒ Δ = 2b²c²(a² + b² – c²) + 2b²c²(c² + a² – b²)

⇒ Δ = 2b²c² [(a² + b² – c²) + (c² + a² – b²)]

⇒ Δ = 2b²c²[2a²]

∴ Δ = 4a²b²c²

Thus,