$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ Prove that
$$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$$
Is true for $x$, $y$ and $z$ being positive integers, with $x$ and $y$ being co-prime and $3z<x<y$.
There are 2 cases in this question:
Case 1 - $x$ is odd and $y$ is even (or $x$ is even and $y$ is odd), in this case $z$ is odd.
Case 2 - Both $x$ and $y$ are odd, in this case $z$ is even.
I tried to show some contradiction, we know that for the statement on the RHS to be true, the prime factors must be present in multiples of 3's, however I struggled going any further than that. 
Any help would be much appreciated.
 A: It is often a proven method to look for coprime factors.
Suppose $q \neq 3$ is a prime dividing $3z-x$. Then $q$ divides $(3z)^3$, hence $z$. This implies that $q$ divides also $x$. Given the symmetry in $x$ and $y$, $q$ cannot divide $3z-y$ as $x$ and $y$ are coprime. Also $x+y$ is not divisible by $q$ as $y$ isn't. 
If $3$ divides $3z-x$ then $3 \mid x$. Hence $3 \nmid y$ as $x$ and $y$ are coprime, and therefor $3 \nmid x+y$ and $3 \nmid 3z-y$. 
We conclude that $x+y$, $3z-x$ and $3z-y$ are mutually coprime.
As $(3z)^3=27z^3$ is a third power, there exist positive coprime integers $u,v,w$ such that $z=uvw$ and either:


*A:
    
*$x+y=9u^3$
    
*$3z-x=v^3$   
    
*$3z-y=w^3$
or


*B:
    
*$x+y=u^3$
    
*$3z-x=9v^3$   
    
*$3z-y=w^3$
Note that we can skip the case $3z-y=9w^3$ as the Diophantine equation is symmetric in $x$ and $y$.A: We have $6z-x-y=6uvw-9u^3=v^3+w^3$. Hence, $6uwv=9u^3+v^3+w^3$. As in general $\sqrt[3]{x_1x_2x_3} \le \frac{x_1+x_2+x_3}{3}$, we have $6uvw<3\sqrt[3]{9}.uvw \le 9u^3+v^3+w^3$, so there is no solution.
B: Likewise we have $6uvw<3\sqrt[3]{9}.uvw \le u^3+9v^3+w^3$, so there is also no solution.
We conclude that there exists no positive integers $x,y,z$ such that $x$ and $y$ are coprime and fulfill the Diophantine equation $(3z)^3=3(x+y)(3z-x)(3z-y)$. QED
A final remark: $z$ is even. The case that $x$ is even and $y$ is uneven implies that $(3z-x)(3z-y)$ is an even number, hence $(3z)^3$ is even.
A: Substituting $t=3z$ and analyzing the function $f(x,y,t)=t^3-3(x+y)(t-x)(t-y)$ we get:
$$f(x,y,t)=t^3-3(x+y)(t-x)(t-y)=t^3-3(x+y)(t^2-t(x+y)+xy)$$
$$f(x,y,t)=t^3-3t^2(x+y)+3t(x+y)^2-(x+y)^3+(x+y)^3-3(x+y)xy$$
$$f(x,y,t)=(t-x-y)^3+(x+y)(x^2+2xy+y^2-3xy)$$
$$f(x,y,t)=(t-x-y)^3+(x+y)(x^2-xy+y^2)=(t-x-y)^3+x^3+y^3$$
Now define $g(x,y,t)=f(-x,-y,t-x-y)=t^3-x^3-y^3$
Let's asume the function $g$ has an integer solution to $g=0$ and let $a,b,c\in\mathbb{N}$ be such a triplet. It follows that $a^3+b^3=c^3$. This contradicts Fermat's Last Theorem and there can not be such triplet solution!
A: Above equation shown below:
$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$
Assume:
$\left(3z\right)^3= 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ ---(1)
For the special case take,
$(x,y,z)=[3m,3n,3(m+n)]$
substituting value of, $(x,y,z)$ in equation (1) we get:
$m^2-mn+n^2=0$  ----(2)
It is well known that equation (2) 
has no integer solution's. Hence a contradiction. 
Hence our assumption is wrong & there is no 
solution for "OP" question for the special case.
