Given, LHS = RHS = 4a 2 b 2 c 2 LHS = Δ = Taking out common factors a, b and c from C 1 , C 2 and C 3 , we have Δ = abc Applying Elementary Transformations R 2 → R 2 – R 1 and R 3 → R 3 – R 1 Δ = abc R 2 → R 2 + R 1 Δ = abc R 3 → R 3 + R 2 Δ = abc Δ = 2ab 2 c C 2 → C 2 – C 1 Δ = 2ab 2 c Expanding along R 3 we get, Δ = 2ab 2 c [a (c – a) + a (a + c)] = 2ab 2 c [ac – a 2 + a 2 + ac] = 2ab 2 c (2ac) = 4a 2 b 2 c 2 Δ = RHS ∴ LHS = RHS Hence, Proved