<s>[INST] 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?
 [/INST] 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.
</s>