Prove the following identities:

Question

Philoid · Accepted Answer

Taking, a,b,c common from C₁, C₂, C₃ respectively we get,

Applying, C₁→C₁ + C₂ + C₃, we get,

Applying, C₂→C₂ – C₁ and C₃→C₃ – C₁, we get,

Applying, C₁→C₁ + C₂ + C₃, we get,

Taking c, a, b common from C₁, C₂, C₃ respectively, we get,

Applying, R₃→R₃ – R₁, we have

= 2a²b²c²(2)

= 4a²b²c² = R.H.S

Hence, proved.