College Algebra, Chapter 3, 3.6, Section 3.6, Problem 36

Determine the functions $f \circ g, \quad g \circ f, \quad f \circ f$ and $g \circ g$ and their domains if $f(x) = x^3 + 2$ and $g(x) = \sqrt[3]{x}$
For $f \circ g$

$
\begin{equation}
\begin{aligned}
f \circ g &= f(g(x)) && \text{Definition of } f \circ g\\
\\
f \circ g &= \left( \sqrt[3]{x} \right)^3 + 2 && \text{Definition of } g\\
\\
f \circ g &= x + 2 && \text{Definition of } f
\end{aligned}
\end{equation}
$

The domain of the function is $(-\infty, \infty)$

For $g \circ f$

$
\begin{equation}
\begin{aligned}
g \circ f &= g(f(x)) && \text{Definition of } g \circ f\\
\\
g \circ f &= \sqrt[3]{(x^3+2)} && \text{Definition of } g\\
\end{aligned}
\end{equation}
$

We know that if the index is an odd number then the domain of function is $(-\infty,\infty)$

For $f \circ f$

$
\begin{equation}
\begin{aligned}
f \circ f &= f(f(x)) && \text{Definition of } f \circ f\\
\\
f \circ f &= (x^3 + 2)^3 + 2&& \text{Definition of } f\\
\\
f \circ f &= x^9 + 6x^6 + 12x^3 + 8 + 2 && \text{Simplify}\\
\\
f \circ f &= x^9 + 6x^6 + 12x^3 + 10 && \text{Definition of } f
\end{aligned}
\end{equation}
$

The domain of the function is $(-\infty,\infty)$

For $g \circ g$

$
\begin{equation}
\begin{aligned}
g \circ g &= g(g(x)) && \text{Definition of } g \circ g\\
\\
g \circ g &= \sqrt[3]{\sqrt[3]{x}} && \text{Definition of } g\\
\\
g \circ g &= \sqrt[6]{x} && \text{Definition of } g
\end{aligned}
\end{equation}
$

We know that if the index is any even number, the radicand can't have a negative value. So the domain of function is $[0,\infty)$