<s>[INST] Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $x^2\cdot f(x)+f(1-x)=2x-x^4$ $\forall\; x\in \mathbb R$ 
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that: $\;x^2\cdot f(x)+f(1-x)=2x-x^4,\;\forall\; x\in \mathbb  R.$

My solution:
Replace $x$ by $(1-x)$ and by eliminating $f(1-x)$.
I obtained $f(x)=1-x^2$ as provided in my book.
My doubt:
How to check that there is no other function satisfying Above property?
Also, I think $f(x)=0,\;\forall\;x\in \mathbb R$ can be solution too.
Is my thinking correct?
 [/INST] Your solution is incomplete, because the solution is not required to be continuous. Your system of equations
$$
\begin{pmatrix}
x^2 & 1 \\
1 & (1-x)^2 
\end{pmatrix}
\begin{pmatrix}
f(x) \\
f(1-x)
\end{pmatrix}
= 
\begin{pmatrix}
2x-x^4 \\
2(1-x) - (1-x)^4
\end{pmatrix}
$$
has infinitely many solutions if $x=\varphi =\frac{1+\sqrt{5}}{2}$ or
$x=1-\varphi =\frac{1-\sqrt{5}}{2}$. In this case, the matrix is not invertible, and we get
$$
\begin{pmatrix}
f(\varphi) \\
f(1-\varphi)
\end{pmatrix}
=\begin{pmatrix}
-\varphi \\
\varphi-1
\end{pmatrix}
+\lambda 
\begin{pmatrix}
-1 \\
1+\varphi
\end{pmatrix}
$$
So for each $\lambda \in \mathbb{R}$, the following is a valid solution:
$$
f(x)=
\begin{cases}
1-x^2 & \text{ for } x\in\mathbb{R}\setminus\{\varphi, 1-\varphi\} \\
-\varphi -\lambda &  \text{ for } x=\varphi \\
\varphi-1 +\lambda(1+\varphi) &  \text{ for } x=1-\varphi
\end{cases}
$$
</s>