Mongolian National Mathematical Olympiad

For any prime $p$ let us set $a = b = p$. Hence $2pf(p)(f(p) + p)$ is a perfect cube, we have $p \mid f(p)$. Let $f(p) = kp$, $k \in \mathbb{Z}_{\ge 1}$. Setting $a = p$, $b = 1, 2$ we get that
$$\begin{aligned} pf(p)^2 + f(1)^2 + 3p(f(p) + f(1)) &= m^3 \\ pf(p)^2 + 2f(2)^2 + 6p(f(p) + f(2)) &= n^3. \end{aligned}$$
Hence $m, n > \sqrt[3]{k^2}p$ and
$$3\sqrt[3]{k^2}p(\sqrt[3]{k^2}p - 1) + 1 < n^3 - m^3 = 3kp^2 + 3p(2f(2) - f(1)) + 2f(2)^2 - f(1)^2.$$
From this inequality we conclude that if $p$ is sufficiently big then $k = 1$.
Now for any positive integer $c$ and for a sufficiently big prime $p$, we set $a = c$ and $b = p$. This yields that
$$(c + p - 1)^3 < cf(c)^2 + p^3 + 3cp(f(c) + p) < (c + p + 1)^3.$$
and therefore $cf(c)^2 + p^3 + 3cp(f(c) + p) = (c + p)^3$. Hence $f(c) = c$. Thus $f$ is the identity map.