Factorize:
(a-b+c)2+(b-c+a)2+2(a-b+c)(b-c+a)
Given,
(a-b+c)2+(b-c+a)2+2(a-b+c)(b-c+a)
= (a-(b-c))2 + (b-(c-a))2 + 2(a-(b-c)) (a+(b-c))
= a2 - 2a (b-c)+(b-c)2+b2-2b(c-a)+(c-a)2+2(a2-(b-c)2)
= a2-2ab+2ac+(b-c)2+b2-2bc+2ab+(c-a)2+2a2-2(b-c)2
= 3a2+2ac-(b-c)2+b2-2bc+(c-a)2
= 3a2+2ac-b2+2bc-c2+b2-2bc+c2+a2-2ac
= 4a2