In solving the equation $z^7-1=0$, the obvious route is to get the root $z=1$. The next step is to solve : $(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=0$. Now, it is difficult for me to solve, and lack of experience is the key. However, have found out that an approach given is to have $a=z+1/z$ to get $a^3+a^2-2a-1=0$. I want to ask why on what basis this approach is made possible. If there were any geometrical reason also, then would be much better. All I could figure out was that the equation obtained in $a,z$ is quadratic, as : $z^2-az+1=0$, with $a$ being the coefficient of the linear term. Second question is for even power of initial expression, i.e. $z^6-1=0$ it would lead to $(z-1)(z^5+z^4+z^3+z^2+z+1)=0$. What can be suitable substitution now? And again what is the algebraic (& if possible also the geometric) reasoning for such substitution.
Also, would request a suitable source to see such substitution text, with either algebraic or geometric reasoning. I hope that Chrystal should serve for algebraic part, but could not find there.