Question 131

If a + b+c= 0, then the value $$(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$$ $$(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$$ is:

Solution

Given : $$a+b+c = 0$$

Let $$a = 1 , b = 1$$ and $$c = -2$$ [We can take any values that satisfy above equation]

To find : $$(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$$ $$(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$$

= $$(\frac{2}{-2} + \frac{-1}{1} + \frac{-1}{1}) (\frac{1}{-1} + \frac{1}{-1} + \frac{-2}{2})$$

= $$(-3) (-3) = 9$$

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