An algebra problem by Ahmad Hesham

$x\quad =\quad 1\quad +\quad { \omega }^{ x }\quad +{ \omega }^{ 2x }\quad \rightarrow \quad (1)\\ \\ By\quad mutiplying\quad by\quad { \omega }^{ x }\quad :\\ \\ x{ \omega }^{ x }\quad =\quad { \omega }^{ x }\quad +\quad { \omega }^{ 2x }\quad +\quad { \omega }^{ 3x }\\ \\ \because { \omega }^{ 3x }=1\\ \\ \therefore x{ \omega }^{ x }\quad =\quad { \omega }^{ x }\quad +\quad { \omega }^{ 2x }\quad +\quad 1\quad \rightarrow \quad (2)\\ \\ From\quad (1)\quad ,\quad (2)\quad :\\ \\ x\quad =\quad x{ \omega }^{ x }\\ \\ x\quad -\quad x{ \omega }^{ x }\quad =\quad 0\\ \\ x(1\quad -\quad { \omega }^{ x })\quad =\quad 0\\ \\ \therefore \quad x\quad =\quad 0\quad \quad Or\quad { \omega }^{ x }\quad =\quad 1\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \therefore \quad x\quad =\quad 3\\ \\ Now\quad there's\quad something\quad I\quad missed\quad which\quad one\quad of\quad \\ the\quad solvers\quad notify\quad me\quad about\quad :)\\ \quad \quad \\ \quad *If\quad you\quad tried\quad it\quad ,\quad you'd\quad find\quad out\quad that\quad \{ x=0\} \quad \\ doesn't\quad satisfy\quad the\quad equation\quad \\ because\quad if\quad you\quad substitute\quad x\quad with\quad 0\quad you'll\quad have\quad :\\ 0=1+1+1\\ 0=3\\ which\quad is\quad rejected\quad of\quad course\\ \\ So\quad we'd\quad come\quad up\quad with\quad \boxed { x=3 } \quad as\quad a\quad final\quad solution$