Question

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

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

Answer 1

Answer: (A)

$-(a+b+c)(a-b)(b-c)(c-a)$

Let

$Q(a,b,c)=a^3(b-c)+b^3(c-a)+c^3(a-b)$

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 $1$ $R(a,b,c)=k(a+b+c),$ where $k=$ constant

$Q(a,b,c)=(a-b)(b-c)(c-a)k(a+b+c)$~~~~~-(1)$

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

The coefficient of $a^{3}b$ in Q(a,b,c) is

so $a\times b\times (-a)\times ka=a^{3}b$

so k=-1

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