Let sixth root of unity be x
So [tex] x^{6}=1 [/tex]
[tex] x^6-1=0 [/tex]
[tex] (x^3)^2-1^2=0 [/tex]
[tex] (x^3-1)(x^3+1)=0 [/tex][tex] x^3-1=0 $ or $ x^3+1=0 [/tex]
[tex] (x-1)(x^2+x+1)=0 $ or $ (x+1)(x^2-x+1)=0 [/tex][tex] x-1=0 $ or $ x^2+x+1=0 $ or $ x+1=0 $ or $ x^2-x+1=0 [/tex]
Using Quadratic formula for the quadratic equations we get
[tex] x=1 $ or $ x=\frac{-1\pm\sqrt{1-4} }{2} $ or $ x = -1 $ or $ x=\frac{1\pm\sqrt{1-4} }{2} [/tex]
[tex] x = 1 $ or $ x=\frac{-1\pm\sqrt{3}i }{2} $ or $ x = -1 $ or $ x=\frac{1\pm\sqrt{3}i }{2} [/tex]
Only 1, out of the given option is the sixth root of unity.
