Prove the following identities:

,

R.H.S = 2(a + b + c)²

Applying C₁→C₁ + C₂ + C₃, we have

Taking, 2(a + b + c) common we get,

Now, applying R₂→R₂ – R₁ and R₃→R₃ – R₁, we get,

Thus, we have

L.H.S = 2(a + b + c)[1(a + b + c)²]

= 2(a + b + c)³ = R.H.S

Hence, proved.