Prove that, there doesn't exist a triple $(x,y,z)\in\Bbb Z_{>0}^3$, such that $$3x^2y+3xy^2+6xyz=z^3$$
My first thought is $z=3k$, this gives $$x^2y+xy^2+6xyk=9k^3$$
So, $$yx^2+x(y^2+6yk)-9k^3=0$$
$$\Delta=(y^2+6yk)^2+36yk^3$$
$$y^4+12y^3k+36y^2k^2+36yk^3$$ should be a perfect square. But this doesn't help. Because of $xy(x+y+6k)=9k^3$, so $xy\mid 9k^3$ but I don't know how to use this. I thought maybe the substitutions $x+y=a, xy=b$ can help, but it doesn't seem so.
The expressions $3x^2y,\;3xy^2$ and $6xyz$ gave me the following observation:
Can we construct an algebraic connection between the expressions
$$(x+y)^3,\; (y+z)^3,\; (x+z)^3,\; (x+y+z)^3\;?$$
Later, I was able to observe that it works very simply:
$$ \begin{align}&(x+y)^3+(y+z)^3-(x+y+z)^3\\ =\;&y^3-3x^2z-3xz^2-6xyz \end{align} $$
Now, letting $x\mapsto x, y\mapsto z, z\mapsto y$ we can obtain:
$$ \begin{align}&3x^2y+3xy^2+6xyz=z^3\\ \iff&(x+z)^3+(y+z)^3=(x+y+z)^3 \end{align} $$
Thus, the thoroughly analyzed Fermat's last theorem for the case $n=3$ will complete the proof.