<s>[INST] Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F(x)+f(-x)=(1+x)^3$ (1). 
\begin{align*}
\left(\frac{F(x)-F(-x)}{e^x}\right)' &=\frac{(f(x)+f(-x))-(F(x)+F(-x))}{e^x}\\ &=\left(\frac{F(x)}{e^x}\right)'-\left(\frac{F(-x)}{e^x}\right)'.
\end{align*}
From here on however I don't know how to solve it. I feel that the two relations from (1) or similar ones can be used.
 [/INST] So, we're solving for differentiable $g$ such that $g(-x) + g'(x) = (1 - x)^3$. Note that $g'(x) = (1 - x)^3 - g(-x)$, and hence $g$ is twice differentiable with
$$g''(x) = -3(1 - x)^2 + g'(-x) = -3(1 - x)^2 + (1 + x)^3 - g(x)$$
That is, $g$ must satisfy the differential equation
$$y'' + y = -3(1 - x)^2 + (1 + x)^3.$$
This differential equation has the general solution
$$y = A \cos(x) + B \sin(x) + x^3 + 3x - 2.$$
But, of course, you're looking for $f = g'$, so the possible solutions we have are
$$f(x) = -A \sin(x) + B \cos(x) + 3x^2 + 3,$$
with corresponding primitive as above.
Do all such functions work? We have
\begin{align*}f(x) + F(-x) &= -A\sin(x) + B\cos(x) + 3x^2 + 3 + A\cos(-x) + B\sin(-x) + (-x)^3 + 3(-x) - 2 \\
&= (1 - x)^3 + (A + B)\cos(x) - (A + B)\sin(x).
\end{align*}
Therefore, we only get a solution with $A = -B$. Thus, the general solution is
$$f(x) = A(\sin(x) + \cos(x)) + 3x^2 + 3,$$
with corresponding primitive
$$F(x) = A(\sin(x) - \cos(x)) + x^3 + 3x - 2.$$
</s>