Question

Factorize $a^3(b-c)+b^3(c-a)+c^3(a-b)$

Answer 1

$f\left(a\right)=a^3(b-c)+b^3(c-a)+c^3(a-b)$
$f\left(b\right)=b^3(b-c)+b^3(c-b)+c^3(b-b)$
$=b^2(b-c)-b^3(b-c)$
$=0$
$\therefore (a-b)$ is a factor
Since the given expression is cyclic, we get $b-c$, $c-a$ as two more factors.
$\therefore$The given expression can be written as
$a^3(b-c)+b^3(c-a)+c^3(a-b)$
$=k(a-b)(b-c)(c-a)(a+b+c)$
Substituting $a=0,b=-1,c=-2$, we get
$-1\left(2\right)+8\left(1\right)=k\left(1\right)(-3)\left(2\right)$
i,e, $k=-1$
$\therefore a^3(b-c)+b^3(c-a)+c^3(a-b)$
$=-(a-b)(b-c)(c-a)(a+b+c)$