Let $e(x)$ be an even function and let $o(x)$ be an odd function, such that
\[e(x) + o(x) = x^2 + x^3\]
for all $x.$ Let $f(x) = e(x) + o(x).$ Find $f(2).$
From the given equation,
\[f(x) = e(x) + o(x) = x^2 + x^3.\]Then
\begin{align*}
f(x) + x^2 - x^3 &= (e(x) + o(x)) + (x^2 - x^3) \\
&= (e(x) + x^2) + (o(x) - x^3) \\
&= g(x) + h(x),
\end{align*}where $g(x) = e(x) + x^2$ and $h(x) = o(x) - x^3.$ Note that $g(x)$ and $h(x)$ are both even functions.
Furthermore, for any $x,$
\[g(-x) = e(-x) + (-x)^2 = e(x) + x^2 = g(x).\]Similarly,
\[h(-x) = o(-x) - (-x)^3 = -o(x) + x^3 = -h(x).\]Then
\[f(-x) + x^2 - x^3 = g(-x) + h(-x) = g(x) - h(x) = f(x) \quad (*).\]Setting $x = 2$ in the original equation, we get
\[f(2) = 2^2 + 2^3 = 8.\]From $(*)$,
\[f(-2) + 2^2 - 2^3 = f(2),\]so $f(-2) = 8.$
Setting $x = 4$ in the original equation, we get
\[f(4) = 4^2 + 4^3 = 80.\]From $(*)$,
\[f(-4) + 4^2 - 4^3 = f(4),\]so $f(-4) = -80.$
Then $g(-2) = f(-2) + 2^2 = 12,$ and $g(-4) = f(-4) + 4^2 = -80.$ Since $g(x)$ is even, $g(2) = g(-2) = \boxed{12},$ and $h(2) = f(2) - g(2) = -4.$
We are given that
\[e(x) + o(x) = x^2 + x^3\]for all $x.$ Substituting $x = 2,$ we get
\[f(2) = e(2) + o(2) = 2^2 + 2^3 = 4 + 8 = \boxed{12}.\]