Question

Factorize:
$(a-b+c{)}^2+(b-c+a{)}^2+2(a-b+c)(b-c+a)$

Answer 1

Given:
$(a-b+c{)}^2+(b-c+a{)}^2+2(a-b+c)(b-c+a)$

$\therefore (a-b+c{)}^2+(b-c+a{)}^2+2(a-b+c)(b-c+a)$
$=\left[\right(a-b+c)+(b-c+a){]}^2$

$[\because x^2+y^2+2xy=(x+y{)}^2]$

$=[a-b+c+b-c+a{]}^2$
$=(2a{)}^2$
$=4a^2$

Hence, $(a-b+c{)}^2+(b-c+a{)}^2+2(a-b+c)(b-c+a)=4a^2$