Answer
$ 2 \sqrt[6]{4 a b^2}+27b\sqrt{ 2 a}$
Work Step by Step
Given \begin{equation}
\sqrt[6]{4 a b^2}+5 b \sqrt{32 a}+\sqrt[6]{4 a b^2}+7 \sqrt{2 a b^2}.
\end{equation} Collect like terms and take the square roots of perfect squares.
\begin{equation}
\begin{aligned}
\sqrt[6]{4 a b^2}+5 b \sqrt{32 a}+\sqrt[6]{4 a b^2}+7 \sqrt{2 a b^2}&= 2 \sqrt[6]{4 a b^2}+5 b \sqrt{16\cdot 2 a}+7b \sqrt{2 a}\\
&=2 \sqrt[6]{4 a b^2}+5b\left( 4\sqrt{ 2 a} \right)+7b \sqrt{2 a} \\
&=2 \sqrt[6]{4 a b^2}+20b\sqrt{ 2 a}+7b \sqrt{2 a} \\
&= 2 \sqrt[6]{4 a b^2}+27b\sqrt{ 2 a}.
\end{aligned}
\end{equation}
