Answer
$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$,
so the given function is the inverse of f.
Work Step by Step
If a function $f$ is one-to-one, then $f^{-1}$ is the unique function for which
$(f^{-1}\circ f)(x)=f^{-1}(f(x))=x$
and $(f\circ f^{-1})(x)=f(f^{-1}(x))=x.$
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$(f^{-1}\circ f)(x)=f^{-1}(f(x))=\sqrt[3]{f(x)+5}$
$=\sqrt[3]{x^{3}-5+5}$
$=\sqrt[3]{x^{3}}$
$=x$
$(f\circ f^{-1})(x)=f(f^{-1}(x))=[f^{-1}(x)]^{3}-5$
$=[\sqrt[3]{x+5}]^{3}-5$
$=x+5-5$
$=x$
$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$,
so the given function is the inverse of f.
