$$\begin{array}{rll}
f(x)&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
y&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
\mbox{The inverse function will be}\\\\
x&=&9(\frac{\sqrt{3}y^4-8}{5}+10)\\\\
\frac{x}{9}&=&\frac{\sqrt{3}y^4-8}{5}+10\\\\
\frac{x}{9}-10&=&\frac{\sqrt{3}y^4-8}{5}\\\\
5\left(\frac{x}{9}-10\right)&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-50&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-42&=&\sqrt{3}y^4\\\\
\frac{5x-42*9}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9\sqrt3}&=&y^4\\\\
\frac{(5x-378)\sqrt3}{27}&=&y^4\\\\
y^4&=&\frac{(5x-378)\sqrt3}{27}\\\\
y&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\
f^{-1}(x)&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\
\end{array}$$
That's what I get anyway.
$$\begin{array}{rll}
f(x)&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
y&=&9(\frac{\sqrt{3}x^4-8}{5}+10)\\\\
\mbox{The inverse function will be}\\\\
x&=&9(\frac{\sqrt{3}y^4-8}{5}+10)\\\\
\frac{x}{9}&=&\frac{\sqrt{3}y^4-8}{5}+10\\\\
\frac{x}{9}-10&=&\frac{\sqrt{3}y^4-8}{5}\\\\
5\left(\frac{x}{9}-10\right)&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-50&=&\sqrt{3}y^4-8\\\\
\frac{5x}{9}-42&=&\sqrt{3}y^4\\\\
\frac{5x-42*9}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9}&=&\sqrt{3}y^4\\\\
\frac{5x-378}{9\sqrt3}&=&y^4\\\\
\frac{(5x-378)\sqrt3}{27}&=&y^4\\\\
y^4&=&\frac{(5x-378)\sqrt3}{27}\\\\
y&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\
f^{-1}(x)&=&\pm\sqrt[4]{\frac{(5x-378)\sqrt3}{27}}\\\\
\end{array}$$
That's what I get anyway.
