Answer
$\frac {p^{15}}{q^{9}}$
Work Step by Step
Given expression : $p(p^{-7}q^{3})^{-2}q^{-3}$
Now, $(ab)^{n}=a^{n}b^{n}$
Hence, $(p^{-7}q^{3})^{-2} = (p^{-7})^{-2}$ X $(q^{3})^{-2}=p^{14}$ X $q^{-6}$
The expression becomes : $p$ X $p^{14}$ X $q^{-6}$ X $q^{-3}=p^{1+14}$ X $q^{-6-3}$
(since the exponents are added while multipliying with the same base)
This becomes : $p^{15}$ X $q^{-9}$
Final result : $\frac {p^{15}}{q^{9}}$ (since $a^{-n}=\frac{1}{a^{n}}$)
