Question Number 194786 by tri26112004 last updated on 15/Jul/23
$${x}^{{n}} +{y}^{{n}} =¿\:\left({n}\in{N}^{\ast} \right) \\ $$
Answered by Frix last updated on 15/Jul/23
$$\mathrm{Let}\:{y}={px} \\ $$$${x}^{{n}} +{y}^{{n}} ={x}^{{n}} \left({p}^{{n}} +\mathrm{1}\right)={x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({p}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right) \\ $$$$\mathrm{Let}\:{p}=\frac{{y}}{{x}} \\ $$$${x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({p}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right)={x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{{y}}{{x}}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right) \\ $$