因数分解解説

$$\begin{array}{rl}{x}^{28}+x^{21}+x^{14}+x^7+1& =\frac{x^{35}-1}{x^7-1}\\ & =\frac{{\displaystyle \prod _{k=0}^{34}(x-e^{\frac{2k\pi }{35}})}}{{\displaystyle \prod _{k=0}^6(x-e^{\frac{2k\pi }{7}})}}\\ & =(x-e^{i\frac{2\pi }{35}})(x-e^{i\frac{4\pi }{35}})(x-e^{i\frac{6\pi }{35}})(x-e^{i\frac{8\pi }{35}})(x-e^{i\frac{12\pi }{35}})\\ & (x-e^{i\frac{14\pi }{35}})(x-e^{i\frac{16\pi }{35}})(x-e^{i\frac{18\pi }{35}})(x-e^{i\frac{22\pi }{35}})(x-e^{i\frac{24\pi }{35}})\\ & (x-e^{i\frac{26\pi }{35}})(x-e^{i\frac{28\pi }{35}})(x-e^{i\frac{32\pi }{35}})(x-e^{i\frac{34\pi }{35}})(x-e^{i\frac{36\pi }{35}})\\ & (x-e^{i\frac{38\pi }{35}})(x-e^{i\frac{42\pi }{35}})(x-e^{i\frac{44\pi }{35}})(x-e^{i\frac{46\pi }{35}})(x-e^{i\frac{48\pi }{35}})\\ & (x-e^{i\frac{52\pi }{35}})(x-e^{i\frac{54\pi }{35}})(x-e^{i\frac{56\pi }{35}})(x-e^{i\frac{58\pi }{35}})(x-e^{i\frac{62\pi }{35}})\\ & (x-e^{i\frac{64\pi }{35}})(x-e^{i\frac{66\pi }{35}})(x-e^{i\frac{68\pi }{35}})\\ & =(x^2-2(\cos \frac{2\pi }{35})x+1)(x^2-2(\cos \frac{4\pi }{35})x+1)(x^2-2(\cos \frac{6\pi }{35})x+1)\\ & (x^2-2(\cos \frac{8\pi }{35})x+1)(x^2-2(\cos \frac{12\pi }{35})x+1)(x^2-2(\cos \frac{14\pi }{35})x+1)\\ & (x^2-2(\cos \frac{16\pi }{35})x+1)(x^2-2(\cos \frac{18\pi }{35})x+1)(x^2-2(\cos \frac{22\pi }{35})x+1)\\ & (x^2-2(\cos \frac{24\pi }{35})x+1)(x^2-2(\cos \frac{26\pi }{35})x+1)(x^2-2(\cos \frac{28\pi }{35})x+1)\\ & (x^2-2(\cos \frac{32\pi }{35})x+1)(x^2-2(\cos \frac{34\pi }{35})x+1)\end{array}$$

$(x^{28}+x^{21}+x^{14}+x^7+1)(x^7-1)=x^{35}-1$に気づくことと,1のn乗根を複素平面でとらえたときに共役な複素数に気づくことができるかどうかが鍵でした。
みゆさん の一般解とフェルマーの最終定理の因数分解のご紹介↓↓↓
みゆさんによる一般解とフェルマーの最終定理の因数分解
ぜひこちらもご覧下さい！