By subtracting both equations: $x = 9y+y^3-6y^2$ and $y = 9x+x^3-6x$ (and assuming x is not equal to y) we get that: $$10 = 6(x+y)-(x^2+xy+y^2)$$Furthermore, if we multiply both equations: Case 1: $(3-x)(3-y)=1$ which means that $8 = 3(x+y)-xy$ Case 2: $(3-x)(3-y)= -1$ which means that $10=3(x+y)-xy$ Let $m = x+y$ and $p = xy$ thus $p = 3m-10$ or $p = 3m-8$ We have two cases: $10 = 9m-m^2-8$ which leaves us with $m=6, p = 10$ or $m=3, p =1$ For which the only x and y which exist are $(\frac{3+\sqrt5}{2}, \frac{3-\sqrt5}{2}$ The second: $10 = 9m-m^2-10$ leaving us with $m = 4, p = 4$ or $m=5, p = 5$ these give us solutions: $(\frac{5+\sqrt5}{2},\frac{5-\sqrt5}{2}$ and $(2+\sqrt2, 2-\sqrt2)$ Finally we take into account $x=y$ which gives us all solutions: $(\frac{5+\sqrt5}{2},\frac{5-\sqrt5)}{2}$, $(2+\sqrt2, 2-\sqrt2)$, $(\frac{3+\sqrt5}{2}, \frac{3-\sqrt5)}{2}$, $(2, 2)$, $(4, 4)$, $(0, 0)$

rafaello wrote: Let $x,y\in\mathbb{R}$ be such that $x = y(3-y)^2$ and $y = x(3-x)^2$. Find all possible values of $x+y$. System is : $y^3-6y^2+9y=x$ $x^3-6x^2+9x=y$ Let $s=x+y$ and $p=xy$ (with $s^2\ge 4p$) Subtracting the two equations, we get $(y-x)(s^2-6s+10-p)=0$ Adding the two equations, we get $s^3-3sp-6s^2+12p+8s=0$ If $x=y$, we get solutions $(0,0),(2,2),(4,4)$ and so $s\in\{0,4,8\}$ If $x\ne y$, first equation implies $p=s^2-6s+10$ Plugging in second, we get $s^3-12s^2+47s-60=0$, with roots $3,4,5$ $s=3$ gives $p=s^2-6s+10=1\le \frac {s^2}4$ and so valid solution $s=4$ already has been seen as a valid solution $s=5$ gives $p=s^2-6s+10=5\le \frac {s^2}4$ and so valid solution Hence the answer $\boxed{x+y\in\{0,3,4,5,8\}}$

Maybe slightly simpler variant of pco's solution: Write $s=x+y, p=xy$. By subtracting the equations, we get $p=s^2-6s+10$ unless $x=y$ (the latter case is easy to deal with). Now by multiplying the equations, we get $p=3s-9 \pm 1$ unless $x=y=0$. Hence $s^2-9s+19 \pm 1=0$ and we just need to solve two quadratic equations for $s$ (instead of a cubic one).

Let $x,y,z\in\mathbb{R}$ be such that $x = y(3-y)^2,y = z(3-z)^2 $ and $z = x(3-x)^2$. Find all possible values of $x+y+z$.

If one of $x,y$ is zero or $x=y$ then we have the solutions $(0,0),(2,2),(4,4)$ giving $0,4,8$ as possible values. From now on, assume $x \neq y, x \neq 0, y \neq 0$. Dividing the two relations, $(x(3-x))^2=(y(3-y))^2$, thus $x(3-x)=y(3-y)$ or $x(3-x)=-y(3-y)$. Case 1: $x(3-x)=y(3-y)$. Then, $(x-y)(x+y-3)=0$, so $x+y=3$ and $x=y(3-y)^2=yx^2$, implying $xy=1$. Since $xy=1 < \frac{9}{4}=\frac{(x+y)^2}{4}$, the values of $x,y$ are indeed real. So $x+y=3$ is another possible value. Case 2: $x(3-x)=-y(3-y)$, so $x^2+y^2=3(x+y)$. By multiplying the two relations, $((3-x)(3-y))^2=1$, so we have two cases. Subcase 1: $(3-x)(3-y)=-1$. Let $x+y=s, xy=p$. Then, $s^2-3s=2p$ and $3s=10+p$, so $s^2-3s=2(3s-10) \Rightarrow (s-4)(s-5)=0$. If $s=4$ then $p=2$ and if $s=5$ then $p=5$, so both value are acceptable. Subcase 2: $(3-x)(3-y)=1$. Then $s^2-3s=2p$ and $3s=8+p$, so $s^2-3s=2(3s-8) \Rightarrow s=\frac{9 \pm \sqrt{17}}{2}$. Note that $s^2-3s=2p \leq s^2/2$, so $0 \leq s \leq 6$, hence $s$ cannot be $\frac{9+\sqrt{17}}{2}$. Thus, $s=\frac{9-\sqrt{17}}{2}$ and $p=\frac{11-3\sqrt{17}}{2}$. However, note that $x=y(3-y)^2=y(\frac{x^2+y^2}{x+y}-y)^2 \Rightarrow xy=(x^2-xy)^2$, so $xy \geq 0$, which is absurd since $11 \leq 3\sqrt{17}$. To sum up, all possible values of $x+y$ are $0,3,4,5,8$.