Find all functions from positive integers to positive integers such that $$f(x + y) = f(x) + f(y) + 3(x + y)\sqrt[3]{f(x)f(y)}$$ for all positive integers $x$ and $y$.
My progress so far: We claim that the unique solution is $$f(x) \equiv x^3.$$ Plugging this in, this works because $$(x + y)^3 = x^3 + y^3 + 3(x + y)xy$$ is true after expanding.
We wish to show that this is the only valid solution. Plug in $(x, x)$ to get $$f(2x) = 2f(x) + 6x\sqrt[3]{f(x)^2}.$$ So, $$\frac{f(2x) - 2f(x)}{6x} = f(x)^{2/3}.$$ The left-hand side is a rational number. Then, since $f(x)$ is an integer, $f(x)^{2/3}$ can only be a rational number if $f(x)$ is a perfect cube. So, we can define a new function $g: \mathbb{N} \to \mathbb{N}$ defined by $g(x) = \sqrt[3]{f(x)}$.
Then, $g$ must satisfy the equation $$g(x + y)^3 = g(x)^3 + g(y)^3 + 3(x + y)g(x)g(y),$$ and we wish to show that $g(x) \equiv x$. Notice that if we can find the value of $g(1)$, then by induction, we will be able to use the above condition to solve for $g$ for all positive integers.
Plug in $(1, 1)$: $$g(2)^3 = 2g(1)^3 + 6g(1)^2.$$ Plug in $(2, 1)$: $$g(3)^3 = g(2)^3 + g(1)^3 + 9g(2)g(1).$$ Plug in $(2, 2)$ and $(3, 1)$: $$g(4)^3 = 2g(2)^3 + 12g(2)^2 = g(3)^3 + g(1)^3 + 12g(3)g(1).$$
Dilemma: This should be solvable to get $g(1) = 1$, but it seems very difficult. Maybe there are some number-theoretic approaches to show that $g(1) = 1$, using the fact that $g$ is defined over the natural numbers?