The inverse of  $f\left(x\right)={\left[4-{\left(x-7\right)}^3\right]}^{\frac{1}{5}}$  is

$x=f^{-1}\left(y\right)$

$y={\left(4-{\left(x-7\right)}^3\right)}^{\frac{1}{5}}$

$\Rightarrow y^5=4-{\left(x-7\right)}^3$

$\Rightarrow 4-y^5={\left(x-7\right)}^3$

$\Rightarrow {\left(4-y^5\right)}^{\frac{1}{3}}=x-7$

$\Rightarrow 7+{\left(4-y^5\right)}^{\frac{1}{3}}=x$

$\Rightarrow f^{-1}\left(y\right)=7+{\left(4-y^5\right)}^{\frac{1}{3}}$

$\Rightarrow f^{-1}\left(x\right)=7+{\left(4-x^5\right)}^{\frac{1}{3}}$