Let Δ = Applying Elementary Transformations C 1 → C 1 + C 2 + C 3 , we have Δ = Taking (a + b + c) common from C 1 , we get Δ = (a + b + c) Applying R 2 → R 2 – R 1 and R 3 → R 3 – R 1 Expanding along C 1 Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0] Δ = (a + b + c) [4bc + 2ab + 2ac + a 2 – a 2 + ac + ba – bc] Δ = (a + b + c) (3ab + 3bc + 3ac) Δ = 3 (a + b + c) (ab + bc + ca) Hence, the given result proved.

Prove that

Let Δ =

Applying Elementary Transformations

C₁→ C₁ + C₂ + C₃, we have

Δ =

Taking (a + b + c) common from C₁, we get

Δ = (a + b + c)

Applying R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Expanding along C₁

Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0]

Δ = (a + b + c) [4bc + 2ab + 2ac + a² – a² + ac + ba – bc]

Δ = (a + b + c) (3ab + 3bc + 3ac)

Δ = 3 (a + b + c) (ab + bc + ca)

Hence, the given result proved.