A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any more?

Once we have proved that all of $x,y,z$ are, if the gcd is 1, cubes themselves, we arrive at, using
$$ \delta = \frac{-1 + i \sqrt 3}{2}  $$
so that $\delta^2 + \delta + 1 = 0$  and $\delta^3=1,$
$$ ( u^3 + v^3 + w^3)^3 - 27  u^3  v^3   w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right)  $$
Let me post a clean proof that, if $\gcd(x,y,z) = 1$  and $(x+y+z)^3 = 27 xyz,$ then actually $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1,$ therefore $x,y,z$ are all cubes. 
We demand, by dividing through if necessary, that $\gcd(x,y,z) = 1.$  Let us ASSUME  that there are a pair that are not coprime, let those be $y,z.$
Next, let $g = \gcd(y,z)$ and ASSUME $g > 1.$ Next we get $y = g \beta$ and $z = g \gamma,$  with $\gcd(\alpha, \beta) = 1.$ The important part is that, from the hypothesis $\gcd(x,y,z) = 1,$ actually
$$ \gcd(x,g) = 1.  $$
Pick some new letters, 
$$ \color{magenta}{mx + ng = 1.} $$
The diophantine equation becomes
$$   27x g^2 \beta \gamma = (x + g \beta + g \gamma)^3 = x^3 + g \cdot (\mbox{stuff})   $$
Thus we have
$$ g | x^3  \; . \; \; $$
Well, $$g | (m x^3 + ng x^2) = x^2 (mx+ng) = x^2$$
so $g | x^2.$
Well, $$g | (m x^2 + ng x) = x (mx+ng) = x$$
so $g | x.$
Well, $$g | (m x + ng ) = 1$$
so $g | 1.$  Then $$ g = 1 $$
which contradicts the assumption that there is a pair of coprime variables among $x,y,z.$ The contradiction proves that $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1.$ As we have $xyz = t^3$ for some integer $t,$ we find that $x,y,z$ are all cubes. Taking $x = u^3, y = v^3, z=w^3,$ the identity 
$$ ( u^3 + v^3 + w^3)^3 - 27  u^3  v^3   w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right)  $$
says that either $x=y=z$ or $x+y+z=0.$
