Since, a + b + c = 0 Then, a + b = -c a + c = -b b + c = a Now, \( \Large \left[ \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \right] \) \( \Large \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \) Now, putting the value of a + b, b + c and c + a from Eqs. (i), (ii) and (iii), we get \( \Large \left[ \frac{ \left(-c\right) }{c} + \frac{ \left(-a\right) }{a} + \frac{ \left(-b\right) }{b} \right] \) \( \Large \left[ \frac{a}{-a} + \frac{b}{-b} + \frac{c}{-c} \right] \) \( \Large \left[ \left(-1\right)+ \left(-1\right)+ \left(-1\right) \left[ \left(-1\right)+ \left(-1\right)+ \left(-1\right) \right] \right] \) \( \Large = \left(-3\right) \times \left(-3\right) = 9 \)