Question

Resolve the following into factors:
(i) ${\left({\sum }_{a,b,c}a\right)}^3-\left({\sum }_{a,b,c}{a}^3\right)$
(ii) ${\sum }_{a,b,c}{a}^2{b}^2(a-b)$

Answer 1

(i) ${\left({\sum }_{a,b,c}a\right)}^3-\left({\sum }_{a,b,c}{a}^3\right)$$f\left(a\right)=(a+b+c{)}^3-(a^3+b^3+c^3)$$f(-b)=(-b+b+c{)}^3-[b^3+b^3+c^3]$$=c^3-c^3=0$
Since the given expression is cyclic, the factors will also be cyclic.
$\therefore (a+b)(c+a)(b+c)$ are three factors. The given expression may have a factor which is a constant
$\therefore$ Let $(a+b+c{)}^3-a^3-b^3-c^3$
$k=(a+b)(b+c)(c+a)$ ......(1)
Put $a=0,b=1,c=2$ in (1) we get,
$(0+1+2{)}^3-0^3-1^3-2^3=k(0+1\left)\right(1+2\left)\right(2+0)$
$\Rightarrow 3^3-1-8\equiv k\left(1\right)k\left(3\right)\left(2\right)$
$\Rightarrow 6k=18ork=3$
$\therefore$ Factors of the given expression are
$(a+b)(b+c)(c+a)$
(ii) ${\sum }_{a,b,c}{a}^2{b}^2(a-b)$
$f\left(a\right)=a^2{b}^2(a-b)+b^2{c}^2(b-c)+c^2{a}^2(c-a)+$
$f\left(b\right)=0+b^2{c}^2(b-c)+c^2{a}^2(c-b)$
$=b^2{c}^2(b-c)-c^2{b}^2(b-c)$
$=0$
$\therefore (a-b)$ is a factor
Since the given expression is cyclic, the factors will also be cyclic.
$\therefore (a-b)(b-c)(c-a)$ are three factors.
The given expression may have a factor which is constant
Let $a^2{b}^2(a-b)+b^2{c}^2(b-c)+c^2{a}^2(c-a)$ .........(1)
put $a=0,b=1,c=2,$ in (1)we get,
$4\left(1\right)=k(-1)(-1)\left(2\right)$
$4=2k$
$k=2$
$\therefore$ Factors of the given expression are
$2(a-b)(b-c)(c-a).$