<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] $\big(x+y+z\big)^{3}=27xyz\tag{1}$
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.
$(x,y,z)=(-(m-n)^3(3m+n)^3(-n^2+6mn+3m^2)^3, 512(3m^2-n^2)^3n^3m^3, (-n+3m)^3(m+n)^3(-n^2-6mn+3m^2)^3).$
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