For any natural $n,m$ prove that
\begin{gather} {\frac { \left|\begin {array}{ccc} {q}^{n+m+2}& \left( {\frac {p}{q}} \right) ^{n+m+2}& \left( {p}^{-1} \right) ^{n+m +2}\\ {q}^{m+1}& \left( {\frac {p}{q}} \right) ^{m+1 }& \left( {p}^{-1} \right) ^{m+1}\\ 1&1&1 \end {array} \right| }{ \left| \begin {array}{ccc} {q}^{2}&{\frac {{p}^{2}}{{q}^{2}}}&{p}^{-2} \\ q&{\frac {p}{q}}&{p}^{-1}\\ 1&1 &1\end {array} \right| }}\\|| \\ {\frac { \left| \begin {array}{ccc} {p}^{n+m+2}& \left( {\frac {q}{p}} \right) ^{n+m+2}& \left( {q}^{-1} \right) ^{n+m +2}\\ {p}^{n+1}& \left( {\frac {q}{p}} \right) ^{n+1 }& \left( {q}^{-1} \right) ^{n+1}\\ 1&1&1 \end {array} \right| }{ \left| \begin {array}{ccc} {p}^{2}&{\frac {{q}^{2}}{{p}^{2}}}&{q}^{-2} \\ p&{\frac {q}{p}}&{q}^{-1}\\ 1&1 &1\end {array} \right| }} \end{gather}
I can prove it by brute force expansion but I hope there are more elegant way to do it.