South African Mathematical Olympiad Third Round

First note that if we replace $f(x)$ by $f(x) + ax + c$ for some constants $a$ and $c$, then we get another solution. Thus we may assume for now that $f(0) = 0$ and $f(1) = 1$.
Let $P(m, n)$ be the statement that was given. Then $P(0, n)$ gives
$$f(n) + f(-n) - 2f(0) = 0$$
and since $f(0) = 0$, we get $f(-n) = -f(n)$. Then taking $P(n, m)$,
$$f(m+n) + f(n-m) - 2f(n) = 6m^2n$$
and subtracting this from $P(m, n)$ gives
$$2f(m-n) + 2f(n) - 2f(m) = 6mn(n-m).$$
Setting $n = -1$, dividing by 2 and rearranging gives
$$f(m+1) = f(m) + f(1) + 3m(m+1).$$
The first few values of $m$ tell us:
$$f(2) = 8, \quad f(3) = 27, \quad f(4) = 64.$$
We hypothesise for an inductive argument that $f(n) = n^3$ and show that
$$f(n+1) = f(1) + f(n) + 3n(n+1) = n^3 + 3n^2 + 3n + 1 = (n+1)^3.$$
Thus $f(n) = n^3$ for all positive integers, and hence by $f(-n) = -f(n)$ also for all negative integers.
Now incorporating our initial remark, we find that all solutions are of the form $f(n) = n^3 + an + c$ for some integer constants $a$ and $c$.