This problem is from an Olympiad handout:
Show that there exists infinitely many integer triplets $(x,y,z)$ such that $ x^3+y^3+z^3-2xyz=1$.
I tried to plug $x=y$ and use tangent lines to find solutions inductively, but the method didn't work well. (It gave $(1,1,1) \rightarrow (13,13,-23)$ ,but after that there was only rational roots)
Also I plugged the equation in the cubic formula and tried to delete the cubic root, but it also failed.
Help me please :I
On
This isn't a full solution, but it's a start. Take modulo $x$, so $y^3+z^3\equiv1$. Set $y^3+z^3=ax+1$. Similarly, set $x^3+z^3=by+1$ and $x^3+y^3=cz+1$. Then $$x^3+y^3+z^3=\tfrac12(ax+by+cz+3).$$ Now express $2xyz$ in terms of $(a,b,c)$ by solving the three simultaneous equations. Namely, $x=\sqrt[3]{\frac12(-ax+by+cz+1)}$ And similarly for the other variables, so $2xyz=\sqrt[3]{(-ax+by+cz+1)(ax-by+cz+1)(ax+by-cz+1)},$ Noting that the factor of $2$ cancels out. Now the expression inside the root must be a perfect cube. Maybe you can try continuing from here?
Above equation shown below,
$x^3+y^3+z^3-2xyz=1$
Along with numerical solutions shown above $(x,y,z)=(1,1,1) = (13,13,-23)$ there is also the solution $ (x,y,z)=( (468/659), (468/659), (-205/659 ) )$