<s>[INST] $\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ 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.
 [/INST] 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!
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