Question

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

Answer 1

${(a + b + c)}^{2} - {(a - b - c)}^{2} + 4 {b}^{2} - 4 {c}^{2} \\ (a + b + c + a - b - c)(a + b + c - a + b + c) + 4( {b}^{2} - {c}^{2} ) \\ 2a(2b + 2c) + 4(b + c)(b - c) \\ 4a(b + c) + 4(b + c)(b - c) \\4( b + c)(a + b - c)$

Answer 2

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)

Step-by-step explanation: