If a + b + c = 0, then the value of \( \Large \left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right) \) \( \Large \left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) \)

A) 9

B) 0

C) 8

D) -3

Correct answer:
A) 9

Description for Correct answer:

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

\( \Large \left[ \frac{ \left(-c\right) }{c} + \frac{ \left(-a\right) }{a} + \frac{ \left(-b\right) }{b} \right] \)

\( \Large \left[ \frac{a}{-a} + \frac{b}{-b} + \frac{c}{-c} \right] \)

\( \Large \left[ \left(-1\right)+ \left(-1\right)+ \left(-1\right) \left[ \left(-1\right)+ \left(-1\right)+ \left(-1\right) \right]  \right] \)

\( \Large = \left(-3\right) \times \left(-3\right) = 9 \)



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