, 2023

By substituting $x = y = 0$ we see that $f(0)^3 = 2f(0)$ which implies $f(0)^3 - 2f(0) = 0$. This means that $f(0) = 0$, $f(0) = -\sqrt{2}$ or $f(0) = \sqrt{2}$.

By substituting $y = 0$ we get $f(x)^3 = f(x^3) + f(0^3) = f(x^3) + f(0)$.

1) Let's first investigate the case $f(0) = 0$: in this case, $f(x)^3 = f(x^3)$, so $(f(x+y))^3 = f(x^3) + f(y^3) = f(x)^3 + f(y)^3$, in which case $g(x) = f(x)^3$ is a Cauchy functional equation and due to function being monotonic must be of the form $c'x$ for some $c' \in \mathbb{R}$. This means that the solution must be $f(x) = \sqrt[3]{c'x} = c\sqrt[3]{x}$. Here $\sqrt[3]{x}$ is a function defined on the entire set of real numbers $\mathbb{R}$. Substituting this to the original equation we get $(c\sqrt[3]{x+y})^3 = c\sqrt[3]{x^3} + c\sqrt[3]{y^3}$ which implies $c^3(x+y) = c(x+y)$, so $c^3 - c = 0$ and $c$ must be $-1$, $0$ or $1$. From this we get solutions $f(x) = 0$, $f(x) = \sqrt[3]{x}$ and $f(x) = -\sqrt[3]{x}$.

2) Let's then investigate the case $f(0) = \sqrt{2}$: By substituting $x = 1$ and $y = 0$ we get $f(1)^3 = f(1^3) + f(0) = f(1) + \sqrt{2}$ and $f(1)^3 - f(1) - \sqrt{2} = 0$. $f(1) = \sqrt{2}$ is a solution, so this can be modified to $(f(1) - \sqrt{2})(f(1)^2 + \sqrt{2}f(1) + 1) = 0$. The latter part doesn't have real roots, so the only option is $f(1) = \sqrt{2}$. Now, substituting $y = 1$ in the original equation we get $f(x+1)^3 = f(x^3) + \sqrt{2}$. For $x = 1$ we get $f(2)^3 = f(1) + \sqrt{2} = 2\sqrt{2}$. This means that also $f(2) = \sqrt{2}$.

Substituting $x = 2$ and $y = 0$ we get $f(2)^3 = f(8) + f(0)$ and substituting $x = 2, y = 1$ we get $f(3)^3 = f(8) + f(1)$, from which we get $f(3) = \sqrt[3]{f(8) + f(1)} = \sqrt[3]{f(8) + \sqrt{2}} = \sqrt[3]{2\sqrt{2} - \sqrt{2} + \sqrt{2}} = \sqrt{2}$. Let us now use induction and assume that $f(x) = \sqrt{2}$ for all $x = 1, 2, \dots, n-1$. Now let's substitute $x = \sqrt[3]{n}, y = 1$: then $f(\sqrt[3]{n} + 1)^3 = f(\sqrt[3]{n}^3) + f(1^3) = f(n) + \sqrt{2}$, from which $f(n) = f(\sqrt[3]{n} + 1)^3 - \sqrt{2} = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$, since $\sqrt[3]{n} + 1 \le n - 1$ when $n \le (n-2)^3$, which is true when $n \ge 4$. Because $f$ is monotonic, also all real values between positive integer values must have $f(x) = \sqrt{2}$.

Let's then substitute $x = -1$ and $y = 0$ in the original equation: we get $f(-1)^3 = f(-1)+f(0) = f(-1)+\sqrt{2}$. From here we can use the same deductions as before to prove that $f(x) = \sqrt{2}$ also for all negative $x$.

3) Let's then investigate the case $f(0) = -\sqrt{2}$: By substituting $x = 1$ and $y = 0$ we get $f(1)^3 = f(1^3) + f(0) = f(1) - \sqrt{2}$ which implies $f(1)^3 - f(1) + \sqrt{2} = 0$. $f(1) = -\sqrt{2}$ is a solution, so this can be modified to $(f(1) + \sqrt{2})(f(1)^2 - \sqrt{2}f(1) + 1) = 0$. The latter part doesn't have real roots, so the only option is $f(1) = -\sqrt{2}$. Now, substituting $y = 1$ in the original equation we get $f(x+1)^3 = f(x^3) - \sqrt{2}$. For $x = 1$ we get $f(2)^3 = f(1) - \sqrt{2} = -2\sqrt{2}$. This means that also $f(2) = -\sqrt{2}$. Proving that $f(3) = -\sqrt{2}$ goes similarly to 2). Similarly to case 2) we can also use induction to first prove that $f(x) = -\sqrt{2}$ then use the similar deductions to 2) to prove it also for all negative $x$.

It's easy to see that constant functions $0$, $\sqrt{2}$ and $-\sqrt{2}$ fulfill the original functional equation. So in the end we get solutions $f(x) = \sqrt[3]{x}$, $f(x) = -\sqrt[3]{x}$, $f(x) = 0$, $f(x) = \sqrt{2}$ and $f(x) = -\sqrt{2}$.