as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.
[tex]\stackrel{h(x)}{y}~~ = ~~6\sqrt[3]{2x+5}-1\implies \stackrel{\textit{quick switcheroo}}{x~~ = ~~6\sqrt[3]{2y+5}-1} \\\\\\ x+1=6\sqrt[3]{2y+5}\implies \cfrac{x+1}{6}=\sqrt[3]{2y+5}\implies \left( \cfrac{x+1}{6} \right)^3=\left( \sqrt[3]{2y+5} \right)^3[/tex]
[tex]\left( \cfrac{x+1}{6} \right)^3=2y+5\implies \left( \cfrac{x+1}{6} \right)^3-5=2y\implies \cfrac{\left( \frac{x+1}{6} \right)^3-5}{2}=y \\\\\\ \cfrac{\left( \frac{x+1}{6} \right)^3}{2}-\cfrac{5}{2}=y\implies \cfrac{~~ \frac{(x+1)^3}{6^3}~~}{2}-\cfrac{5}{2}=y\implies \cfrac{(x+1)^3}{432}-\cfrac{5}{2}=\stackrel{y}{h^{-1}(x)}[/tex]
