1) Observe the trivial solution $x=y=z=0$.

2) Observe that if $x$, $y$, $z$ is a solution, then so is $x\text{'}=-x$, $y\text{'}=-y$, $z\text{'}=-z$.

3) If $x=y=z$ is to be a solution, then

$(x+y{)}^3=z$

$\Rightarrow$

$(2x{)}^3=x$

If $x\ne 0$ [which impiles $x=y=z=0$] then

$x=y=z=1/\sqrt{8}$ [the observation (2) above covering the negative solution]

4) Without loss of generality (and having recorded the solution (1)) we can assume

$x\ge y\ge z>0$. But $x<1$ [else $(x+y{)}^3=z\Rightarrow z>1$, and then that $y>1$. But they can't all be > 1]. So

$1>x\ge y\ge z>0$

5) What if $z>1/\sqrt{8}$? Write $z=\mu /\sqrt{8}$. Then $(x+y{)}^3\ge (2\mu /\sqrt{8}{)}^3=8{\mu }^3/8\sqrt{8}={\mu }^3/\sqrt{8}$. The latter is $>\mu /\sqrt{8}=z$ so the first equation is not satisfied. So $z<1/\sqrt{8}$ [no need for 'less than or equal to' here since equality would imply that they are all equal, which is a solution we already have]

6) Having recorded solutions (1), (3), any reamining solution must have $z<1/\sqrt{8}$, from (5). Can we have $y\ge 1/\sqrt{8}$? No, because if so, $x=\mu /\sqrt{8}$, $y=\beta /\sqrt{8}$ where $\mu$ and $\beta$ are both $\ge 1$. But then, $(x+y{)}^3=(\mu +\beta {)}^3/8\sqrt{8}\ge 1/\sqrt{8}$, which is strictly greater than $z$, so the first equation is not satisfied. So both $y$ and $z$ are $<1/\sqrt{8}$. The second equation implies $x<1/\sqrt{8}$ also. So it remains to see whether there is a solution where

$1/\sqrt{8}>x\ge y\ge z>0$

The answer is no. Write

$x=\alpha /\sqrt{8}$

$y=\beta /\sqrt{8}$

$z=\chi /\sqrt{8}$

Then $(y+z{)}^3=1/8\sqrt{8}(\beta +\chi {)}^3\le 1/8\sqrt{8}(\beta +\beta {)}^3=1/\sqrt{8}{\beta }^3<1/\sqrt{8}\beta$. The latter must be $<x$, therefore the first equation is not satisfied.

So the only solutions are the ones described in (1), (2), (3).