Question

If $a+b+c=0$, then value of $\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}$ is:

• A. $3$
• B. $-3$
• C. $-\frac{1}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $-3$

We know that in general, $a^3+b^3+c^3=3abc$ when $a+b+c=0$.

Since $a+b+c=0$, therefore, $a+b=-c$, $b+c=-a$ and $c+a=-b$.

Now consider, $\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}$

$=\frac{a^2(-a)+b^2(-b)+c^2(-c)}{abc}=\frac{-a^3-b^3-c^3}{abc}$

$=\frac{-1(a^3+b^3+c^3)}{abc}=\frac{-3abc}{abc}=-3$.

Hence, $\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}=-3$.

Therefore, option $B$ is correct.