[tex]\bf f(x)=y=6x^3-8\qquad inverse\implies
\begin{array}{lll}
x=&6y^3-8\\
&\uparrow\\
&\textit{switched variables}
\end{array}
\\\\\\
x+8=6y^3\implies \cfrac{x+8}{6}=y^3\impliedby taking\ \sqrt[3]{\qquad }
\\\\\\
\sqrt[3]{\cfrac{x+8}{6}}=\sqrt[3]{y^3}\implies \sqrt[3]{\cfrac{x+8}{6}}=y\impliedby f^{-1}(x)[/tex]
