<s>[INST] Reason for substitution : $a=z + \frac{1}{z}$. Have read regarding the substitution $a = z + \frac{1}{z}$ to get the factorization of $z^6+z^5+z^4+z^3 +z^2+z+1$ to get the form $a^3+a^2-2a -1=0$ in the book by Erickson martin, titled: Beautiful mathematics, on page #$58$ as shown below.
I am unable to get the process for the division, i.e. how to divide by $a$ the given polynomial. I mean that $a = \frac{z^2+1}{z}$ cannot divide $z^6+z^5+z^4+z^3 +z^2+z+1$.
For $z^5-1=0\implies (z-1)(z^4+z^3+z^2+z+1)=0$, want to use the same logic of symmetry for $z+z^4= z+\frac1z=a'$, but am hindered by the inability to divide $(z^4+z^3+z^2+z+1)$ by $a'$.

 [/INST] I don't understand why you are referring to division.  All you have to do is substitute
$$a=z+\frac1z$$
into
$$a^3+a^2-2a-1$$
and simplify.  You should get
$$\frac{z^6+z^5+z^4+z^3+z^2+z+1}{z^3}\ .$$
(I believe the $z^6+z^5+z^4+z^3+z^2+1$ in your question is wrong.)  Then you have
$$z^6+z^5+z^4+z^3+z^2+z+1=0\quad\hbox{if and only if}\quad a^3+a^2-2a-1=0\ .$$
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