[tex]x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)=[/tex]
[tex](x-1)(x+1)(x-\dfrac{1+i\sqrt3}{2})(x-\dfrac{1-i\sqrt3}{2})(x+\dfrac{1+i\sqrt3}{2})(x+\dfrac{1-i\sqrt3}{2})[/tex]
Am folosit faptul ca solutiile ecuatiei
[tex]x^2-x+1=0\ sunt\ x_{1,2}=\dfrac{1\pm i\sqrt3}{2}[/tex], iar ale ecuatiei
[tex]x^2+x+1=0\ sunt\ x_{1,2}=\dfrac{-1\pm i\sqrt3}{2}[/tex]
