Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z}$ such that $|mn|\ne 1$ and $y=x+m$ and $z=x-n$? Another way would be assuming at least 2 of them are non-consecutive?
I would like to see general ideas on how to proceed.

Use the well-known identity $$x^3+y^3+z^3-3xyz = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right)$$ to have, from your given equation (since the left hand side equals $3(x+y+z)$, $$\begin{aligned} 3(x+y+z) & = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right) \\ \implies 6 &= (x-y)^2+(y-z)^2+(z-x)^2 \ (\because x,y,z>0 \implies  x+y+z\ne 0) \end{aligned}$$
also, $x,y,z$ being positive integers, the squares in the last line $(x-y)^2, (y-z)^2, (z-x)^2$ are perfect square integers, and the only way three perfect squares add up to $6$ is $$6=1+1+4$$
so, two of the differences $|x-y|$ and $|y-z|$ (without loss of generality) must be $1$, naturally the $|x-z|$ must be $2$.
