Question

Factorise for $a=1,b=1,c=1$ in $a^2(b-c)+b^2(c-a)+c^2(a-b)$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We simply substitute $a=1,b=1,c=1$ in the expression $a^2(b-c)+b^2(c-a)+c^2(a-b)$ as follows:

$a^2(b-c)+b^2(c-a)+c^2(a-b)=1^2(1-1)+1^2(1-1)+1^2(1-1)=(1\times 0)+(1\times 0)+(1\times 0)=0+0+0=0$

Hence, $a^2(b-c)+b^2(c-a)+c^2(a-b)=0$ when $a=1,b=1,c=1$.