Value of $f(f(x) – 2y) = 2x – 3y + f(f(y) – x)$ Let $f : \mathbb{R} \rightarrow \mathbb{R} $ be a polynomial function satisfying 
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x),$$
where $x, y \in \mathbb{R}$. Find the value of $$f(21) – f(14).$$ 
 A: You are given
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x) \tag{1}\label{eq1A}$$
Since $x,y \in \mathbb{R}$, you can let $x = \frac{3y}{2}$ in \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
f\left(f\left(\frac{3y}{2}\right) - 2y\right) & = 2\left(\frac{3y}{2}\right) - 3y + f\left(f\left(y\right) - \frac{3y}{2}\right) \\
f\left(f\left(\frac{3y}{2}\right) - 2y\right) & = f\left(f\left(y\right) - \frac{3y}{2}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note $f$ can't be the constant function since \eqref{eq1A} then wouldn't hold for all $x$ and $y$. Actually, as WhatsUp's comment states, $f$ must be linear. To see this, consider the degree of $f$ to be some $n \gt 1$. Then $f(x) - 2y$ has an $x$ term of degree $n$ and $f(f(x) - 2y)$ has an $x$ term of degree $n^2$. However, on the right side, $f(f(y) - x)$ would only have a highest degree term of $x$ being $n$. Since $n^2 \gt n$, it would not be possible for \eqref{eq1A} to always hold.
Since $f$ must be linear, then
$$f(y) = ay + b \tag{3}\label{eq3A}$$
for some real $a \neq 0$ and $b$. Let the arguments to the outer $f$ functions in \eqref{eq2A} be $g(y)$ and $h(y)$ to get using \eqref{eq3A} that $ag(y) + b = ah(y) + b \implies a(g(y) - h(y)) = 0 \implies g(y) = h(y)$. Thus, you get
$$f\left(\frac{3y}{2}\right) - 2y = f(y) - \frac{3y}{2} \tag{4}\label{eq4A}$$
Using $y = 14$ in \eqref{eq4A} gives
$$f(21) - 28 = f(14) - 21 \implies f(21) - f(14) = 7 \tag{5}\label{eq5A}$$
Alternatively, you can use \eqref{eq3A} in \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
f(ax + b - 2y) & = 2x - 3y + f(ay + b - x) \\
a(ax + b - 2y) + b & = 2x - 3y + a(ay + b - x) + b \\
a^2x + ab - 2ay & = 2x - 3y + a^2y + ab - ax \\
a^2x - 2ay & = 2x - 3y + a^2y - ax \\
(a^2 + a - 2)x & = (a^2 + 2a - 3)y 
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
For this to be true for all real $x$ and $y$ requires $a^2 + a - 2 = a^2 + 2a - 3 = 0 \implies a = 1$. Using $a = 1$ in \eqref{eq3A} gives
$$f(21) - f(14) = (21 + b) - (14 + b) = 7 \tag{7}\label{eq7A}$$
This matches what was obtained in \eqref{eq5A}.
A: Here is my Approach:
$f(f(x) – 2y) = 2x – 3y + f(f(y) – x)$
If $x=21,y=14, then: 2x-3y=0$ So
$ f(f(21)-28)=f(f(14)-21)$
$f(21)-f(14)=7$
