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 $a'$.

We say that a polynomial is palindromic, if its sequence of coefficients can equally well be read backwards. So a degree $n$ polynomial $$ p(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n,\ a_n\neq0, $$ is palindromic if $a_i=a_{n-i}$ for all indices $i, 0\le i\le n$. This system of equations can compactly be restated in the form $$ p(x)=x^np(\frac1x). $$ This way of doing it shows that $p(\alpha)=0$ if and only if $p(1/\alpha)=0$.

If we further assume that $n$ is even, say $n=2k$, then we get to the business part. In that case we can write $$ \frac1{x^k}p(x)=a_0x^{-k}+a_1x^{-k+1}+\cdots+a_{k-1}x^{-1}+a_k+a_{k+1}x+\cdots a_{2k}x^k.\qquad(*) $$ Observe that here the coefficients of $x^i$ and $x^{-i}$ are equal as a consequence of the palindromic property. This means that $(*)$ can be written as a polynomial in the new variable $z=x+\dfrac1x$. Behold $$ \begin{aligned} x+\frac1x&=z,\\ x^2+\frac1{x^2}&=(x+\frac1x)^2-2=z^2-2,\\ x^3+\frac1{x^3}&=(x^2+\frac1{x^2})(x+\frac1x)-(x+\frac1x)=z^3-3z,\\ \vdots\\ x^{\ell+1}+\frac1{x^{\ell+1}}&=(x^\ell+\frac1{x^\ell})(x+\frac1x)-(x^{\ell-1}+\frac1{x^{\ell-1}})=\cdots.\\ \end{aligned} $$ Here the last line contains a general recurrence relation defining a sequence of polynomials $q_\ell, \ell=1,2,\ldots$, such that $x^\ell+x^{-\ell}=q_\ell(z)$. We simply declare $q_0(z)=2, q_1(z)=z$, and apply the recurrence $$ q_{\ell+1}(z)=z q_\ell(z)-q_{\ell-1}(z) $$ for higher values of $\ell$.

Therefore the right hand side of $(*)$ is equal to $$a_k+a_{k-1}q_1(z)+a_{k-2}q_2(z)+\cdots+a_0q_k(z)=\sum_{i=0}^ka_{k-i}q_i(z).\qquad(**)$$

Observe that $\deg q_i=i$, so on the right hand side of $(**)$ we have a degree $k$ polynomial.

What all this implies is that we can find the zeros of a palindromic polynomial $p(x)$ of even degree $n=2k$ by the process of:

1. Write $x^k(p(1/x)$ in the form $f(z)$ with a polynomial $f$ of degree $k$.
2. Find the zeros of $f(z)$ (this may still be taxing if $k$ is large).
3. For each zero $z_j, j=1,2,\ldots,k,$ solve the quadratic equation $$x+\frac1x=z_j$$ to find two of the zeros of $p(x)$.

Example. When $p(x)=x^4+x^3+x^2+x+1$, a palindromic of degree $4=2\cdot2$, we see that $$ x^2p(\frac1x)=x^2+x+1+\frac1x+\frac1{x^2}=1+q_1(z)+q_2(z)=z^2+z-1. $$ The zeros of $z^2+z-1$ are $z_{1,2}=(-1\pm\sqrt5)/2$. The rest is easy.

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 $$z^6+z^5+z^4+z^3+z^2+z+1=0\quad\hbox{if and only if}\quad a^3+a^2-2a-1=0\ .$$