Ans: $\dfrac{\beta^{23}}{\alpha^9}$
$\begin{array}{cl}
& (\alpha \beta)^3(\alpha^{-2}\beta^4)^5 \\
= & \alpha^2 \beta ^3 \times \alpha^{-10}\beta^{20} \\
= & \alpha^{-8}\beta^{23} \\
= & \dfrac{\beta^{23}}{\alpha^8}
\end{array}$
$\begin{array}{cl}
& (\alpha \beta)^3(\alpha^{-2}\beta^4)^5 \\
= & \alpha^2 \beta ^3 \times \alpha^{-10}\beta^{20} \\
= & \alpha^{-8}\beta^{23} \\
= & \dfrac{\beta^{23}}{\alpha^8}
\end{array}$
