Question Number 46881 by maxmathsup by imad last updated on 02/Nov/18
![factorize inside C[x] the polynom x^n +y^n](https://www.tinkutara.com/question/Q46881.png)
$${factorize}\:{inside}\:{C}\left[{x}\right]\:{the}\:{polynom}\:\:{x}^{{n}} \:+{y}^{{n}} \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
![let cnsider the polynome p(x)=x^n +y^n if y=0 p(x)=x^n if y≠0 let find roots of p(x) p(x)=0 ⇔x^n =−y^n ⇔((x/y))^n =−1 =e^(i(2k+1π) ⇒(x/(y )) = e^(i(((2k+1)π)/n)) ⇒ the roots of p(x) are x_k =y e^(i(((2k+1)π)/n)) with k ∈[[0,n−1]] ⇒ p(x) =λ Π_(k=0) ^(n−1) (x−x_k ) but λ =1 ⇒p(x) =Π_(k=0) ^(n−1) (x−y e^(i(((2k+1)π)/n)) ) =(x−ye^(i(π/n)) )(x−y e^(i((3π)/n)) ).....(x−y e^(i(((2n−1)π)/n)) ) .](https://www.tinkutara.com/question/Q46914.png)
$${let}\:{cnsider}\:{the}\:{polynome}\:{p}\left({x}\right)={x}^{{n}} \:+{y}^{{n}} \:\:{if}\:{y}=\mathrm{0}\:\:\:{p}\left({x}\right)={x}^{{n}} \\ $$$${if}\:{y}\neq\mathrm{0}\:\:{let}\:{find}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow{x}^{{n}} \:=−{y}^{{n}} \:\Leftrightarrow\left(\frac{{x}}{{y}}\right)^{{n}} \:=−\mathrm{1}\:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\pi\right.} \:\Rightarrow\frac{{x}}{{y}\:}\:=\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \:\Rightarrow \\ $$$${the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{x}_{{k}} ={y}\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \:\:{with}\:{k}\:\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\Rightarrow \\ $$$${p}\left({x}\right)\:=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{x}_{{k}} \right)\:\:{but}\:\lambda\:=\mathrm{1}\:\Rightarrow{p}\left({x}\right)\:=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{y}\:{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \right) \\ $$$$=\left({x}−{ye}^{{i}\frac{\pi}{{n}}} \right)\left({x}−{y}\:{e}^{{i}\frac{\mathrm{3}\pi}{{n}}} \right)…..\left({x}−{y}\:{e}^{{i}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{{n}}} \right)\:. \\ $$