Answer :
Given, 3a2 – b2 – c2 + 2ab – 2bc + 2ca
⇒ 4a2 – a2 – b2 – c2 + 2ab – 2bc + 2ca
⇒ (2a)2 – (a2 + b2 + c2 – 2ab + 2bc – 2ca)
⇒ (2a)2 – (a – b – c)2
[Since, from the identity III we know that, (x2 – y2) = (x – y)(x + y)]
⇒ (2a – (a – b – c))(2a + a + b + c)
⇒ (a + b + c)(3a + b + c)
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