Consider the monic cubic $f(x)$ with roots $\dfrac{a}{b},\dfrac{b}{c}, \dfrac{c}{a}$. We claim $f(x)$ is an integer polynomial. Of course, by vieta's formulas, it suffices to show that \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}, \] \[ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} ,\] and \[\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}\] are integers. The first two are given to us; the last is equal to $1$. By the rational root theorem then, $\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\in \{-1, 1\}$. Thus $|a|=|b|=|c|$ and the possible sums are $3$ and $3$, or $-1$ and $-1$ respectively.

By the rational root theorem then, $\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\in \{-1, 1\}$ Why is this so?

An alternative take also using the rational root theorem is that (assuming the polynomial $f(x)$ is monic, that it has roots $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ and that it is an integer polynomial) the denominators of the roots must divide the leading coefficient, in this case $1$. Therefore, $|a|=|b|=|c|=1$, which easily gives the sums $(3,3)$ and $(-1,-1)$