Question

Simplify and factorise: (a+b+c) 2 -(a-b-c) 2 +4b 2 -4c 2

Answer 1

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

Answer:

= 4(b + c)(a + b – c)

Step-by-step explanation:

let (a+b+c)2 – (a – b – c)2+ 4b2– 4c2

= a2 + b2 + c2 + 2ab + 2bc + 2ca – (a2 + b2 + c2 – 2ab + 2bc – 2ca)+ 4b2– 4c2

= a2 + b2 + c2 + 2ab + 2bc + 2ca – a2 – b2 – c2 + 2ab – 2bc +2ca + 4b2– 4c2

= 4ab + 4ca + 4b2– 4c2

= 4(ab + ca + b2– c2)

= 4[a(b + c) +(b2– c2)]

= 4[a(b + c) + (b + c)(b – c)]

= 4(b + c)(a + b – c)

(answer)