Question Number 29308 by tawa tawa last updated on 07/Feb/18
$$\mathrm{Solve}:\:\:\:\mathrm{w}^{\mathrm{3}} \:=\:−\:\mathrm{16} \\ $$
Answered by Tinkutara last updated on 07/Feb/18
$$\frac{{w}^{\mathrm{3}} }{−\mathrm{16}}=\mathrm{1} \\ $$$$\frac{{w}}{−\mathrm{2}^{\mathrm{4}/\mathrm{3}} }=\mathrm{1},\omega,\omega^{\mathrm{2}} \\ $$$${w}=−\mathrm{2}^{\mathrm{4}/\mathrm{3}} ,−\mathrm{2}^{\mathrm{4}/\mathrm{3}} \omega,−\mathrm{2}^{\mathrm{4}/\mathrm{3}} \omega^{\mathrm{2}} \\ $$
Answered by sma3l2996 last updated on 07/Feb/18
$${w}^{\mathrm{3}} =−\mathrm{16}=\mathrm{16}{i}^{\mathrm{2}} =\mathrm{16}{e}^{{i}\pi} \\ $$$${w}_{{k}} =\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{3}}\right)\pi} \\ $$$${w}_{\mathrm{0}} =\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{{i}\pi/\mathrm{3}} \:\:;\:{w}_{\mathrm{1}} =−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\:;\:\:{w}_{\mathrm{2}} =\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{−{i}\pi/\mathrm{3}} \\ $$