y = x³ + 5 . Solve this equation for x:
x³ = y - 5 ---> (x³)^1/3 = (y - 5)^1/3 ---> x = (y - 5)^1/3 , now type y instead of x, and x instead of y
                                                       y = (x - 5)^1/3 
  in term of f(x) ---->                   f^-1(x) = (x - 5)^1/3  

f^-1(x) denotes the inverse, not 1/f(x),

Let x  =  (f^-1(x))^3  + 5

x  -  5  =  (f^-1(x))^3

f^-1(x) = (x - 5)^(1/3)