8-8. The numbers $a, b$, and $c$ (not necessarily integers) are such that

$$
a+b+c=0 \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=100
$$

What is $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ ?
Answer: -103.

Solution. We have

$$
100=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{-b-c}{b}+\frac{-a-c}{c}+\frac{-a-b}{a}=-3-\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)
$$

from which $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}=-103$.

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