If a+b+c=0, find the value of (a+b)/c-2b/(c+a)+(b+c)/a. (a) 0 (b) 1 (c) -1 (d) 2

Given the equation \(a + b + c = 0\), we can solve for one of the variables in terms of the others. Without loss of generality, let’s solve for \(c\):

\[
c = -a – b
\]

Substituting this into the given expression:

\[
\frac{a+b}{c} – \frac{2b}{c+a} + \frac{b+c}{a} = \frac{a+b}{-a-b} – \frac{2b}{-a-b+a} + \frac{b-a-b}{a}
\]

This simplifies further to:

\[
\frac{a+b}{-a-b} – \frac{2b}{-b} + \frac{-a}{a}
\]

Since \(\frac{a+b}{-a-b} = -1\) (because the numerator is the negation of the denominator), and \(\frac{-a}{a} = -1\), the expression further simplifies to:

\[
-1 + \frac{2b}{b} – 1 = -1 + 2 – 1
\]

\[
= 2 – 2 = 0
\]

Thus, the value of the given expression, under the condition that \(a + b + c = 0\), is \(0\).

Answer: (a) 0.