Question

Resolve the following into factors:

$\sum _{a,b,c}{a}^2{b}^2(a-b)$

• A. $-2(a-b)(b-c)(c-a)$
• B. $2(a-b)(b-c)(c-a)$
• C. $2(a-b)(b-c)(c+a)$
• D. None

Answer 1

The correct option is B $2(a-b)(b-c)(c-a)$

Let

Q(a,b,c)=$\sum a^2{b}^2(a-b)$

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

If we interchange the value of a ,b ,c then we will get the same polynomial

Q(a,b,c)=Q(b,c,a)=Q(c,a,b)

So a-b is a factor of Q(b,b,c)

As the polynomial is cyclic so

b-c and c-a are also the factor of Q

The only symmetrical polynomial of degree 1 is $k$, where k=constant

Q(a,b,c)=(a-b)(b-c)(c-a)k $-\left(1\right)$

To find the value of k compare the coefficient of one terms.

we get k=2

(1)$\to Q(a,b,c)=2(a-b)(b-c)(c-a)$