$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ 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.

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.
