Suppose $$f(x) = \frac{x}{8}-3 \quad \hbox{and} \quad g(x) =x^3.$$ Then

$(f\circ g)(x) =$ .
$(f\circ g)^{-1}(x) =$ .
$(f^{-1}\circ g^{-1})(x) =$ .
$(g^{-1}\circ f^{-1})(x) =$ .
(You can enter a cube root by raising x to the one-third power. Don't forget the parentheses around the exponent.)