Question Number 216487 by Tawa11 last updated on 08/Feb/25
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\omega^{\mathrm{7}} \:\:+\:\:\omega^{\mathrm{8}} \:\:+\:\:\omega^{\mathrm{12}} \:\:\mathrm{where} \\ $$$$\omega\:\:\mathrm{is}\:\mathrm{omega}\:\mathrm{function}. \\ $$
Commented by mr W last updated on 09/Feb/25
$${can}\:{you}\:{specify}\:{what}\:{your}\:{omega} \\ $$$${function}\:\omega\:{means}? \\ $$
Commented by Tawa11 last updated on 09/Feb/25
$$\mathrm{Ohh}. \\ $$$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{sir}. \\ $$$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{not}\:\mathrm{specifying}. \\ $$
Answered by Rasheed.Sindhi last updated on 09/Feb/25
$$\left.\begin{matrix}{\omega^{\mathrm{3}} =\mathrm{1}}\\{\omega^{\mathrm{2}} +\omega+\mathrm{1}=\mathrm{0}}\end{matrix}\right\} \\ $$$$=\left(\omega^{\mathrm{3}} \right)^{\mathrm{2}} .\omega+\left(\omega^{\mathrm{3}} \right)^{\mathrm{2}} .\omega^{\mathrm{2}} +\left(\omega^{\mathrm{3}} \right)^{\mathrm{4}} \\ $$$$=\omega+\omega^{\mathrm{2}} +\mathrm{1} \\ $$$$=\mathrm{0} \\ $$
Commented by Tawa11 last updated on 09/Feb/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$