If a + b + c = 0, then $\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac}$ is equal to :

1. 0

2. 1

3. 3

4. abc

Given,

a + b + c = 0

⇒ a + b = -c

⇒ b + c = -a

⇒ c + a = -b

Substituting the above values in $\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac}$, we get :

$\Rightarrow \dfrac{(-c)^2}{ab} + \dfrac{(-a)^2}{bc} + \dfrac{(-b)^2}{ca} \\[1em] \Rightarrow \dfrac{c^2}{ab} + \dfrac{a^2}{bc} + \dfrac{b^2}{ca} \\[1em] \Rightarrow \dfrac{a^3 + b^3 + c^3}{abc} \text{ ……(1)}$

We know that,

If, a + b + c = 0 then a³ + b³ + c³ = 3abc

Substituting the value of a³ + b³ + c³ in (1), we get :

$\Rightarrow \dfrac{3abc}{abc} \\[1em] \Rightarrow 3.$

Hence, option 3 is correct option.