+0

# 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 +1 68 1 +588 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(\frac1a + \frac1c\right) + c\left(\frac1a+\frac1b\right).$

Sep 1, 2022

#1
+39
0

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.

Sep 3, 2022