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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 $c^{\prime }x$ for some $c^{\prime }\in \mathbb{R}$. This means that the solution must be $f(x)=\sqrt[3]{c^{\prime }x}=c\sqrt[3]{x}$. 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 $f(x)=-\sqrt{2}$.