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 $a'$.

 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:


*

*Write $x^k(p(1/x)$ in the form $f(z)$ with a polynomial $f$ of degree $k$.

*Find the zeros of $f(z)$ (this may still be taxing if $k$ is large).

*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.
A: $a=z+\dfrac{1}{z}\Rightarrow a^2=\left(z+\dfrac{1}{z}\right)^2=z^2+\dfrac{1}{z^2}+2$
$a=z+\dfrac{1}{z}\Rightarrow a^3=\left(z+\dfrac{1}{z}\right)^3=z^3+3\dfrac{z^2}{z}+3\dfrac{z}{z^2}+\dfrac{1}{z^3}=z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}$
$a=z+\dfrac{1}{z}\Rightarrow -2a=-2z-\dfrac{2}{z}$
Then we will have
$a^3+a^2-2a-1=0$
$\Leftrightarrow z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}+z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}-1=0$
$\Leftrightarrow z^3+x^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}=0$
$\Leftrightarrow z^3\left(z^3+x^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}\right)=0$
$\Leftrightarrow z^6+z^5+z^4+z^3+z^2+z+1=0$
If this problem appears as an exercise, it should be done like this (reverse the steps above):

$z^6+z^5+z^4+z^3+z^2+z+1=0$, we know that $z=0$ is not a root of this equation
$\Leftrightarrow \dfrac{z^6+z^5+z^4+z^3+z^2+z+1}{z^3}=0$
$\Leftrightarrow z^3+z^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}=0$
$\Leftrightarrow z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}+z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}-1=0$
Let $a=z+\dfrac{1}{z}$, we have:
$a=z+\dfrac{1}{z}\Rightarrow a^2=\left(z+\dfrac{1}{z}\right)^2=z^2+\dfrac{1}{z^2}+2$
$a=z+\dfrac{1}{z}\Rightarrow a^3=\left(z+\dfrac{1}{z}\right)^3=z^3+3\dfrac{z^2}{z}+3\dfrac{z}{z^2}+\dfrac{1}{z^3}=z^3+\dfrac{1}{z^3}+3z+\dfrac{3}{z}$
$a=z+\dfrac{1}{z}\Rightarrow -2a=-2z-\dfrac{2}{z}$
then $a^3+a^2-2a-1=0$.

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