Question

How to factorise (a+b+c)2+(b-c+a)2+2(a-b+c)(b-c+a)

Answer 1

Answer:

here we use the identity x^2 - y^2 =(x+y)(x-y)

(a+b+c)2-(a-b-c)2+4(b2-c2)

= [a+b+c+a-b-c][a+b+c-(a-b-c)]+4(b+c)(b-c)

= 2a*[a+b+c-a+b+c]+4(b+c)(b-c)

=2a*[2b+2c]+4(b+c)(b-c)

=2a*2(b+c)+4(b+c)(b-c)

=4(b+c)[a+(b-c)] taking 4(b+c) common

=4(b+c)(a+b-c)

Answer 2

Hope it's Helps u.... ...............