Question

Factorize : $a^2(b+c{)}^2+b^2(c+a{)}^2+c^2(a+b)+abc(a+b+c)+(a^2+b^2+c^2)(bc+ca+ab)$

Answer 1

$f\left(a\right)=a^2(b+c{)}^2+b^2(c+a{)}^2+c^2(a+b{)}^2+abc(a+b+c)+(a^2+b^2+c^2\left)\right(bc+ca+ab)$
$f(-b)=b^2(b+c{)}^2+b^2(c-b{)}^2-b^2{c}^2+(2b^2+c^2)(bc-bc-b^2)$
$=b^2(b^2+c^2+2b^4{c}^2+b^2-2bc)$
$=b^2[2b^2+2c^2]$
$=2b^4+|2b^2{c}^2-b^2{c}^2-2b^4-b^2{c}^2=0$
$\therefore (a+b)$ is a factor
Since the expression is cyclic,
$(a+b)(b+c)(c+a)$ are factors.
$\therefore k(a+b+c)(a+b)(b+c)(c+a)$
By putting $a=0,b=1,c=2,$ we get
$k=1$
$\therefore (a+b+c)(a+b)(b+c)(c+a)$ are the factors.