If f(x)

We have,

$f(x)=4-(x-7)^{3}$

Let $y=4-(x-7)^{3}$

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

$\Rightarrow x-7=\sqrt[3]{4-y}$

$\Rightarrow x=7+\sqrt[3]{4-y}$

$\Rightarrow f^{-1}(y)=7+\sqrt[3]{4-y}$

$\therefore f^{-1}(x)=7+\sqrt[3]{4-x}$