Answer
$-\dfrac{a^8}{2b^5}$
Work Step by Step
RECALL:
(i) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$
Divide the numerical coefficients then use rule (i) above to obtain:
$=\frac{-10}{20} \cdot a^{5-(-3)}b^{6-11}
\\=-\frac{1}{2} \cdot a^{5+3}b^{-5}
\\=-\frac{1}{2} a^8b^{-5}$
Use rule (ii) above to obtain:
$=-\dfrac{1}{2} \cdot a^8 \cdot \dfrac{1}{b^5}
\\=-\dfrac{a^8}{2b^5}$
