#### Answer

$\dfrac{t^{12}}{r}$

#### Work Step by Step

RECALL:
(i) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne0$
(ii) $a^0=1, a\ne0$
(iii) $a^{-m} = \dfrac{1}{a^m}. a\ne0$
Use the rule above to have:
$=r^{2-3}s^{4-4}t^{6-(-6)}
\\=r^{-1}s^{0}t^{6+6}
\\=\dfrac{1}{r} \cdot 1 \cdot t^{12}
\\=\dfrac{t^{12}}{r}$