If $$a + b + c = 0$$ then the value of $$\frac{1}{(a + b)(b + c)} + \frac{1}{(b + c)(c + a)} + \frac{1}{(c + a)(a + b)}$$ is

$$a + b + c = 0$$ 

 $$\frac{1}{(a + b)(b + c)} + \frac{1}{(b + c)(c + a)} + \frac{1}{(c + a)(a + b)}$$

taking LCM we get 

 $$\frac{c+a}{(a + b)(b + c)(c+a)} + \frac{(a+b)}{(a+b)(b + c)(c + a)} + \frac{(b+c)}{(c + a)(b+c)(a + b)}$$

= $$\frac{c+a+a+b+b+c}{(a + b)(b + c)(c+a)}$$ = $$\frac{2(a+b+c)}{(a + b)(b + c)(c+a)}$$

we know $$a + b + c = 0$$ 

$$\frac{c+a+a+b+b+c}{(a + b)(b + c)(c+a)}$$ = $$\frac{2(a+b+c)}{(a + b)(b + c)(c+a)}$$ = 0