$Consider \ x^7+1,$ Using the formula, $a^n+b^n = (a+b) (a^{n-1}-b a^{n-2}+b^2 a^{n-3}+ \cdots + b^{n-1})$ $x^7+1^7 = (x+1) (x^6 -x^5 \times 1 + x^4 \times 1^2 - x^3 \times 1^3+ x^2 \times 1^4 - x \times 1^5 +1^6) , $ $\therefore \ x^7+1=0 (x+1)(x^6-x^5+x^4-x^3+x^2 -x+1) $ $\to x+1=0, \ x^6-x^5+x^4-x^3+x^2-x+1 =0 $ Hence solving $x^7+1=0 $ and inoring the root $x=-1,$ Gives all the roots of $x^6-x^5+x^4-x^3+x^2-x+1$ $x^7=-1 = \cos (\pi +2\pi k)+i \sin (\pi + 2\pi k), \ k=0 \ to \ 6$ $x= [\cos (\pi + 2 \pi k)+ i \sin (\pi +2\pi k)]^{\frac {1}{7}} = \left [ \cos \dfrac {\pi + 2 \pi k}{7} + i \sin \dfrac {\pi +2\pi k}{7} \right ] $ $when \ k=0, $ $x =\cos \dfrac {\pi}{7} + i \sin \dfrac{\pi}{7} \ \cdots (A)$ $ \ when \ k=1, $ $\ x=\cos \dfrac {3 \pi}{7}+ i \sin \dfrac {3 \pi}{7} \ \cdots (B) $ $when \ k=2, $ $x=\cos \dfrac {5 \pi}{7}+ i \sin \dfrac {5 \pi}{7} \ \cdots (C)$ $when \ k=3, $ $ x=\cos \pi + i \sin \pi = -1 \ \cdots ( ignored) $ $ when \ k=4, $ $x=\cos \dfrac {9 \pi}{7}+ i \sin \dfrac {9 \pi}{7} \ \cdots (D) $ $ when \ k=5, $ $x=\cos \dfrac {11 \pi}{7}+ i \sin \dfrac {11 \pi}{7} \ \cdots (E)$ $when \ k=6 , $ $x=\cos \dfrac {13 \pi}{7}+ i \sin \dfrac {13 \pi}{7} \ \cdots (F)$ values (A) through (F) are the 6 roots of $x^6-x^5+x^4-x^3+x^2-x+1=0$ $x=\left [ \cos \left ( \dfrac {\pi+2\pi k}{7} \right ) + i \sin \left ( \dfrac {\pi + 2\pi k} {7} \right ) \right ] , \ k=0,1,2,4,5,6$