Find $f(x)$ satisfying the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ A function $f(x)$ satisfies the  functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ for all real $x$. Then $f(x)$ is given by.
My work 
$$x^2{f(x)} +f(1- x) =2x -x^4$$
Replacing $x$ $by$ $1- x$ 
$$(1-x)^2{f(1 - x)} +f(x) =2x -x^4 + 4x^3 -6x^2 + 1$$
Subtracting them
$${f(x)}(1-x^2) +{f(1- x)}((1-x)^2 - 1) = 4x^3 -6x^2 + 1$$
What should I do next ?
 A: You have the equation 
\begin{equation}
x^2\cdot f(x) + f(1-x) = 2x-x^4.
\end{equation}
Replacing $x$ by $1-x$ results in
\begin{equation}
(1-x)^2\cdot f(1-x) + f(x) = 2(1-x)-(1-x)^4.
\end{equation}
Now we have two equations with two unknowns, namely $f(x)$ and $f(1-x)$. We can subtract $(1-x)^2$ times the first equation from the second one to get
\begin{equation}
\left(1-(1-x)^2x^2\right)\cdot f(x) = 2(1-x)-(1-x)^4 - (1-x)^2\cdot (2x-x^4).
\end{equation}
This can be simplified to $$f(x)=1-x^2.$$
A: If we assume (the reasoning given in comments) $f(x)$ to be a quadratic $ax^2 + bx + x$, then put $x = 0, \frac12, 1$ in the given equation, you get 3 equations.
$f(0) = 1 = c$
$f(1) = 0 = a + b + c$ and
$f(\frac{1}{2}) = \frac{3}{4} = \frac{a}{4} + \frac{b}{2} + c$
Solving, you get $a = -1$ and $c = 1$ so the function is $f(x) = \color{blue}{1 - x^2}$
A: It is easier to solve a much more general problem:
Consider the functional equation
a(x) f(x) + b(x) f (1-x) = g(x).
Call a^* (x) = a(1-x) etc. and the equation becomes the two equations.
a f + b f^* = g
a^* f^* + b^* f = g^*. which is a 2-by-2 linear system:
Solving for f by inverting a 2-by-2 matrix gives
f(x) = (1/ (a a^* - b b^)) ( a^ g - b^* g^*).
This works for all functions a,b,g as long as a a^* - b b^* is not zero.
The given problem has a(x) = x^2, b(x) = 1, and g(x) = 2x-x^4.
Plugging these in gives
f(x) = (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 g(x) - g(1-x) )
 = (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 (2x-x^4) - 2(1-x) + (1-x)^4) 

which simplifies to 1-x^2.
