Given that,
$\large q^{-a} = \dfrac{1}{r},\;r^{-b}= \dfrac{1}{s},\; s^{-c} =\dfrac{1}{q}$
Now, $\large q^{-a} = \dfrac{1}{r}$
$\large\Rightarrow \dfrac{1}{q^{a}} =\dfrac{1}{r}$
$\large\Rightarrow q^{a} = r$
Similarly, $\large r^{b} = s$ and $\large s^{c} = q$
In $\large q^{a} = r$, we replace $q$ by $\large s^{c}$ and $r$ by $\large s^{\dfrac{1}{b}}$
We get, $\large s^{ca} = s^{\dfrac{1}{b}}$
Since, the bases are equal, we can work with the exponents.
So, we get, $ac$ = $\dfrac{1}{b}$ $\Rightarrow abc = 1$