Answer
$f^{-1}(x)=(x-3)^3+4$
Work Step by Step
$f(x)=\sqrt[3] {x-4}+3$
$y=\sqrt[3] {x-4}+3$
a.
$x=\sqrt[3] {y-4}+3$
$x-3=\sqrt[3] {y-4}$
$(x-3)^3=y-4$
$f^{-1}(x)=(x-3)^3+4$
b.
$f(f^{-1}(x)=\sqrt[3] {(x-3)^3+4-4} +3=x-3+3=x$
and
$f^{-1}(f(x))=(\sqrt[3] {x-4} +3-3)^3+4=x-4+4=x$
