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?
 A: $\big(x+y+z\big)^{3}=27xyz\tag{1}$
Let $s=x+y+z$ and substitute $z=s-x-y$ to equation $(1).$
$s^3-27xys+27x^2y+27xy^2=0\tag{2}$
For $x$ to be rational solution, the discriminant of the equation $(2)$ must be square.
Hence $-27y(4s-3y)(s-3y)^2$ must be square, then
$v^2 = (9y-6s)^2-36s^2$
We solve simultaneous equations $[6s=m^2-n^2,9y-6s=m^2+n^2].$
We get $y = \frac{2m^2}{9}, s = \frac{m^2-n^2}{6}, x = \frac{-(m-n)^3}{36m}, z=\frac{-(m+n)^3}{36m}.$
Finally, $(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3).$
$(m,n)=(2,1): [x,y,z]=[-1, 64, -27]$
$(m,n)=(3,2): [x,y,z]=[-1, 216, -125]$
$(m,n)=(4,1): [x,y,z]=[-27, 512, -125]$
$(m,n)=(4,3): [x,y,z]=[-1, 512, -343]$ 
Added new solutions:
We can get new solution using a known solution of equation $(1).$
$(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3)$ is a known solution.
Substitute $x = t-(m-n)^3, y = t+8m^3, z = kt-(m+n)^3$ to equation $(1)$,
and let $k= \frac{-(5m^4+8m^3n+2m^2n^2+n^4)}{4(m^2-2mn+n^2)m^2},$ then $t = \frac{-72(m^4-2m^3n+2n^3m-n^4)m^3}{9m^4-18m^3n-6n^3m-n^4}.$
Hence we get new parametric solution below.
$(x,y,z)=(-(3m+n)^3(m-n)^3, -64n^3m^3, (m+n)^3(-n+3m)^3).$ 
Similarly, we get other new solution using above new solution below.
$(x,y,z)=(-(m-n)^3(3m+n)^3(-n^2+6mn+3m^2)^3, 512(3m^2-n^2)^3n^3m^3, (-n+3m)^3(m+n)^3(-n^2-6mn+3m^2)^3).$
A: 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 $x+y+z=0.$
A: We have the identity:
$a^3+b^3+c^3=3abc$
Condition is, $(a+b+c)=0$
In "OP" solution if we take,
$a=(r+s)(r-3s)$
$b=2r(2s-r)$
$c=(r^2-2rs+3s^2)$
We get, $(a+b+c)=0$
Hence, $a^3+b^3+c^3=3abc$  ----(1)
Therefore by cubing both sides of equation (1),
$(a^3+b^3+c^3)^3=(3abc)^3=27(abc)^3 =27(a^3)(b^3)(c^3)$
Let, $(x,y,z)=(a^3,b^3,c^3)$
And we get:
$(x+y+z)^3=27xyz$
The constant, $w=(r^4-4r^3s+4r^2s^2+3s^4)$ 
is a multiplication factor & can be multiplied in if required.
