Problem based on Algebraic Identities and Factorisation

Question

I was preparing for an entrance exam and found a question a sample paper that I could not solve.

Given, $$ a + b + c = 0 $$ Find the value of $$ a^2 (b + c) + b^2 (c + a) + c^2 (a + b) \over abc $$ I tried to do this many different ways. Some of my findings are - $$ a^2 (b + c) + b^2 (c + a) + c^2 (a + b) \over abc $$ $$= {ab + ac \over bc} + {bc+ab \over ac} + {ac+bc \over ab} $$ $$ = {(ab)^2 +(ac)^2 +2a^2 bc+(ab)^2 +(bc)^2 +2ab^2 c+(bc)^2 +(ac)^2+2abc^2\over (ab)(ac)(bc)}$$ $$ = {(ab +ac)^2 +(ab+bc)^2 +(bc+ac)^2\over (ab)(ac)(bc)}$$ Am I headed the right way?

Answer 1

plugging $$c=-a-b$$ in the given term $$\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)}{abc}$$ and simplifying we get $$-3$$ as the searched result.

Answer 2

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

Since $a+b+c=0$ we have:

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

And knowing that

$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc] \Rightarrow$$

$$ 0=a^3 + b^3 + c^3-3abc \Rightarrow a^3 + b^3 + c^3=3abc$$

Backing to $(*)$ we get:

$$-{a^3 + b^3 + c^3 \over abc}=-{3abc \over abc}=-3$$