<s>[INST] What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
 [/INST] There's an algebraic way (which may be a bit messy without using a CAS).  It involves exploiting the symmetry of the problem, as Peter Tamaroff suggested.
$$
\begin{eqnarray}
  x^2 f(x) + f(1-x) &=& 2x - x^4 \\
  (1-x)^2f(1-x) + f(x) &=& 2(1-x) - (1-x)^4 \quad \textrm{subs. $1-x$ for $x$}\\
  \\
  -(1-x)^2x^2f(x) - (1-x)^2f(1-x) &=& -(1-x)^2(2x - x^4) \quad \textrm{Eqn.1 times $-(1-x)^2$}\\
  \textrm{Now add to Eqn.2:} \quad f(x) - (1-x)^2x^2f(x) &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\\
  f(x)\left[1 - (1-x)^2x^2\right] &=& 2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)\quad \textrm{factoring out $f(x)$}\\
  f(x) &=& \frac{2(1-x)-(1-x)^4 - (1-x)^2(2x-x^4)}{1 - (1-x)^2x^2}.
\end{eqnarray}
$$
This is an unreduced final answer, which (according to sage) cancels out to leave $f(x) = 1 - x^2$.
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