This is problem 719 in the exercises book by Faddeev and Sominski. Here is
their proof.
Denote by $D$ the set of all real univariate polynomials all of whose roots are
real.
Lemma 1 If $f\in D$ and $\lambda \in {\mathbb R}$, then
$f'+\lambda f \in D$ also.
Proof of lemma 1. The case $\lambda=0$ follows from Rolle’s theorem, so assume $\lambda \neq 0$. Denote by $x_1<x_2< \ldots <x_r$ the distinct roots of
$f$, and denote by $m_i$ the multiplicity of $x_i$ in $f$. Then $\sum_{k=1}^r m_k=n$ where
$n$ is the degree of $f$. Consider the rational fraction $g=\frac{f'}{f}$. Its poles are
the $x_k$, and $g$ is surjective $(x_k,x_{k+1}) \to {\mathbb R}$ for each $k\lt n$. Also, on the border, $g$ is surjective $(-\infty,x_1) \to (-\infty,0)$ and surjective $(x_n,+\infty) \to (0,+\infty)$. We deduce that there is a $y_k\in (x_k,x_{k+1})$ with $g(y_k)=-\lambda$ and that there is an yet another $y$ satisfying $g(y)=-\lambda$ (so $y$ will be $\lt x_1$ if $\lambda$ is negative and $\gt x_n$ if $\lambda$ is positive). Let us now count all the roots we have found for $f'+\lambda f$ : we have $y,y_1,y_2, \ldots y_{r-1}$, plus the $x_k$ with multiplicities $m_k-1$. This makes up a total of $n$ real
roots and concludes the proof.
Lemma 2 If $f,g\in D$ with $g=\sum_{k=0}^n {\gamma}_k x^k$, then
$h=\sum_{k=0}^n {\gamma}_k f^{(k)} \in D$ also.
Proof of lemma 2. Let $\lambda_1, \ldots ,\lambda_n$ denote the roots
of $g(-x)$, so that $g(x)=\prod_{k=1}^n (x+\lambda_k)$. Iterating lemma 1 above, we see
successively that $F_1=f’+\lambda_1f, F_2=F_1'+\lambda_2 F_1, \ldots$, etc, up to
$F_{n}=F_{n-1}'+\lambda_n F_{n-1}=h$, are all in $D$.
Lemma 3 Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$, and $m$ be a positive integer. Then $\sum_{k=0}^n \frac{m!}{(m-n+k)!} a_k x^k \in D$ also.
Proof of lemma 3. Use lemma 2 with $g(x)=x^m$, and multiply by $x^{n-m}$ if needed.
Main theorem Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$. Then
$\sum_{k=0}^n \binom{n}{k} a_k x^k \in D$ also.
Proof of main theorem. Using lemma 3 with $m=n$, we see that
$h_1=\sum_{k=0}^n \frac{n!}{k!} a_k x^k \in D$. So $x^nh_1(\frac{1}{x}) \in D$ also,
which means that $h_2=\sum_{k=0}^n \frac{n!}{(n-k)!} a_k x^k \in D$. Using lemma 3 again
with $h_2$ in place of $f$ and $m=n$, we see that
$h_3=\sum_{k=0}^n \frac{n!}{k!} \frac{n!}{(n-k)!} a_{n-k} x^k \in D$. Then, divide by
$n!$.
There are several ways to do it. One straightforward way is to factor the octic into two quartics over a square root extension, then use the quartic formula. Quartics have a cubic resolvent, but since the $17$th root of unity is construcible, then we know this cubic will also factor over a square root.
I. Factoring the octic
Given,
$$x^8+x^7-7x^6-6x^5+15x^4+10x^3-10x^2-4x+1= 0$$
Factor this into two quartics over $\sqrt{17}$ choosing,
$$x^4 + \frac{1 - \sqrt{17}}2x^3 - \frac{3 + \sqrt{17}}2x^2 + (2 + \sqrt{17})x - 1 = 0$$
such that $x = 2\cos \left(\frac{2\pi}{17} \right)$. Depress the quartic with the transformation,
$$x = y+\frac{-1 + \sqrt{17}}8$$
to get,
$$y^4 + \frac{-51 - 5\sqrt{17}}{16}y^2 + \frac{17 + 7\sqrt{17}}{16}y + \frac{221 + 43 \sqrt{17}}{512} = 0$$
II. Depressed quartic
Given,
$$y^4+py^2+qy+r = 0$$
then its cubic resolvent is simply,
$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$
With any non-zero root $u$, then the depressed quartic's formula for the four roots are,
$$y_k = \frac12\left(\sqrt{u}\pm\sqrt{-(2p+u)-\frac{2q}{\sqrt{u}}}\right)$$
for $\pm\sqrt{u}$ and you are done.
III. Cubic resolvent
Given $(p,q,r)$ we find the resolvent as,
$$u^3 + \frac{-51 - 5 \sqrt{17}}8u^2 + \frac{323 + 53 \sqrt{17}}{32}u - \left(\frac{17 + 7\sqrt{17}}{16}\right)^2 = 0$$
this has the root,
$$u = \frac{17-\sqrt{17}}{8}$$
Using the formula and reversing the transformation, we find,
$$x = \frac{-1+\sqrt{17}}8+\frac12\sqrt{u}+\frac12\sqrt{\sqrt{17}+2u-\sqrt{\frac{17}{u}}-\sqrt{u}}$$
After some minor algebraic manipulation, we end up with the original form,
$$\small{x_1 = 2\cos \left(\frac{2\pi}{17} \right) = \frac{1}{8} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)}
+ 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right)}$$
IV. Conjugate
Using one of its conjugates,
$$\small{x_2 = 2\cos \left(\frac{8\pi}{17} \right) = \frac{1}{8} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)}
\color{red}{-} 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right)}$$
then,
$$t=x_1+x_2 = 2\cos \left(\frac{2\pi}{17} \right)+2\cos \left(\frac{8\pi}{17} \right) = \frac{1}{4} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)}\right)$$
a root of,
$$t^4 + t^3 - 6t^2 - t + 1 = 0$$
which was the second question of the OP.
Best Answer
Remove the first two terms with $\left(z^2+\frac12z\right)^2$: $$ z^4+z^3+z^2+2z+3-\left(z^2+\frac12z\right)^2=\frac34z^2+2z+3\tag1 $$ Remove the first two terms of $\frac34z^2+2z+3$ with $\frac34\left(z+\frac43\right)^2$: $$ \frac34z^2+2z+3-\frac34\left(z+\frac43\right)^2=\frac53\tag2 $$ Thus, we get $$ z^4+z^3+z^2+2z+3=\left(z^2+\frac12z\right)^2+\frac34\left(z+\frac43\right)^2+\frac53\tag3 $$