Question

Resolve the following into factors:
(i) ${\sum }_{a,b,c}a(b^4-c^4)$ (ii) ${\sum }_{a,b,c}(a+1{)}^2(b-c)$

Answer 1

(i) $f\left(a\right)=a(b^4-c^4)+b(c^4-a^4)+c(a^4-b^4)$
$f\left(b\right)=0$
$\therefore (a-b)(b-c)(c-a)$ are factors given expression being cyclic.
The given expression can be written as
$a(b^4-c^4+b(c^4-a^4)+c(a^4-b^4)$
$\equiv k(a-b)(b-c)(c-a)$
$(a^2+b^2+c^2+ab+bc+ca)$......(1)
Put $a=0;b=1;c=2$ in (1),
$1\left(16\right)+2(-1)\equiv k(-1)(-1)\left(2\right)(1+4+2)$
$14=14k$
$k=1$
$\therefore$ Factors of the given expression are
$(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)$
(ii) ${\sum }_{a,b,c}(a+1{)}^2.(b-c)$
$f\left(a\right)=(a+1{)}^2.(b-c)+(b+1{)}^2.(c-a)+(c+1{)}^2.(a-b)$
$f\left(b\right)=(b+1{)}^2.(b-c)+(b+1{)}^2.(c-b)+(c+1{)}^2.(b-b)$
$=(b+1{)}^2.(b-c)+(b+1{)}^2.(b-c)=0$
$\therefore (a-b)$ is a factor of the given expression.
Since the given expression is cyclic, the factors are also cyclic.
$\therefore (b-c),(c-a)$ will be also factors.
The given expression is of third order.
The product of the factors is also of third order. If at all there is any factor, it will be constant. Let it be 'K'.
$\therefore$ The given expression can be written as,
$(a+1{)}^2.(b-c)+(b+1{)}^{c-a}+(c+1{)}^2(a-b)$
$\equiv K(a-b)(b-c)(c-a)$ .......(1)
This being an identiy we can give any values to a, b, c in (1) we get the value of K.
Put $a=0;b=1;c=2$
$(0+1{)}^2.(1-2)+(1+1{)}^2.(2-0)+(2+1{)}^2.(0-1)$
$\equiv K(0-1)(1-2)(2-0)$
$\equiv K(-1)(-1)\left(2\right)$
$\Rightarrow -1+8-9=K\left(2\right)$
$2K=-2$ $\therefore K=\frac{-2}{2}=-1$
$\therefore$ Factors of the given expression
$=-\left[\right(a-b\left)\right(b-c\left)\right(c-a\left)\right]$