45th Mongolian Mathematical Olympiad

This equation is same as $x^3-8={\Phi }_{17}(y)$, ${\Phi }_{17}(y)$ is 17th cyclotomic polynomial. $p$ is prime number, if $p|{\Phi }_{17}(x)$ then $p\equiv 1(\mathrm{mod}17)$ or $p|17$. So if for arbitrary $d|{\Phi }_{17}(y)$ then $d\equiv 1(\mathrm{mod}17)$ or $d|17$. $(x-2)(x^2+2x+4)={\Phi }_{17}(y)$. If we have for $d=x-2|{\Phi }_{17}(y)$ then $x-2\equiv 1(\mathrm{mod}17)\iff x\equiv 3(\mathrm{mod}17)$.

If we have $17|x-2$ then $x\equiv 2(17)$; $$x^2+2x+4\equiv 9+6+4\equiv 2(17).$$ But $x^2+2x+4$ polynomial is ${\Phi }_{17}(y)$ polynomial's divisor then contradict to above two cases.