Let $a$, $b$, and $c$, be nonzero real numbers such that $a+b+c=0$. Compute the value of \[ a\left(\frac 1b + \frac1c\right) + b\left(\frac1

Expanding and simplifying, we get \(\frac{a+c}b+\frac{a+b}c+\frac{b+c}a\).

 

Since \(a+b+c=0\), we also have \(a+c=-b, \)

\(c+b=-a,\) and \(a+b=-c.\)

 

Substituing into the simplified expression, we have \(-1+(-1)+(-1)\), which is just \(\boxed{-3.}\) Very simple problem with nice solution.