3a² + 4ab + b² – 2ac – c²

Given,3a² + 4ab + b² – 2ac – c²

⇒ 3a² + 4ab + b² – 2ac – c²

⇒ 4a² + 4ab + b² – a² – 2ac – c²

⇒ From the identity I, a² + 2ab + b² = (a + b)²

⇒ 4a² + 4ab + b² – (a² + 2ac + c²)

⇒ (2a + b)² – (a + c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((2a + b) – (a + c))

((2a + b) + (a + c)

⇒ (2a + b – a – c)(2a + b + a + c)