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 ?
On
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 $f(x) = \color{blue}{1 - x^2}$
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 $$f(x)=1-x^2.$$