If a+b+c=0 then find value of $$\frac{(b + c)^2}{bc} + \frac{(c + a)^2}{ca} + \frac{(a + b)^2}{ab}$$ ? ​

Correct option is D

Given:

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a + b + c = 0

Formula used:

​$$a + b + c = 0, \text{then } a^3 + b^3 + c^3 = 3abc\\\ \\\text{Thus, } a^3 + b^3 + c^3 = 3abc$$​​

Solution:

$$a + b + c = 0\\\text{Then } a + b = -c, \, b + c = -a \, \text{and} \, c + a = -b\\\frac{(b + c)^2}{bc} + \frac{(c + a)^2}{ca} + \frac{(a + b)^2}{ab}\\\ \\= \frac{bc}{(-a)^2} + \frac{ca}{(-b)^2} + \frac{ab}{(-c)^2}\\\ \\= \frac{bc}{a^3} + \frac{ca}{b^3} + \frac{ab}{c^3}\\\ \\\text{As we know, if } a + b + c = 0, \text{then } a^3 + b^3 + c^3 = 3abc\\\ \\\text{Thus, } a^3 + b^3 + c^3 = 3abc\\\ \\\frac{abc}{3abc} = \frac{3abc}{abc} = 3$$​​