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$\begin{aligned} x^2 - x - 1 &= 0\\x^2 &=x+1\\x &=1+\frac{1}{x}...(1)\\x+\frac{1}{x}&=1+\frac{2}{x}\\\frac{x^{16} - 1}{x^8 + 2x^7} &=\frac{x^{8} - \frac{1}{x^8}}{1 + \frac{2}{x}}\\ &=\frac{x^{8} - \frac{1}{x^8}}{x+\frac{1}{x}}\\ &= \color{#D61F06}{\left( x^{4} +\frac{1}{x^4} \right)}\color{#3D99F6}{\left( x^{2} +\frac{1}{x^2} \right)}\color{#20A900}{\left( x -\frac{1}{x} \right)} \end{aligned}$
From $(1)$

$\begin{aligned} x-\frac{1}{x} &=\color{#20A900}{1}\\x^2+\frac{1}{x^2}-2 &=1\\x^2+\frac{1}{x^2} &=\color{#3D99F6}{3}\\x^4+\frac{1}{x^4} &=\color{#D61F06}{7}\end{aligned}$

Substituting the values thus obtained,

$\begin{aligned}\frac{x^{16} - 1}{x^8 + 2x^7} &=\color{#D61F06}{7}.\color{#3D99F6}{3}.\color{#20A900}{1}\\ &\Huge\color{#3D99F6}{=\boxed{21}} \end{aligned}$

Suppose $\alpha$ is a root of $x^2-x-1=0$ Then $\alpha^2=\alpha+1$ $\alpha^4=(\alpha+1)^2=\alpha^2+2\alpha+1=(\alpha+1)+2\alpha+1=3\alpha+2\\$ $\alpha^8=(\alpha^4)^2=(3\alpha+2)^2=9\alpha^2+12\alpha+4=9(\alpha+1)+12\alpha+4=21\alpha+13\\$ $\alpha^{16}=(21\alpha+13)^2=441\alpha^2+169+546\alpha=441(\alpha+1)+546\alpha+169=987\alpha+610\\$ $\alpha^7=\alpha \cdot \alpha^2 \cdot \alpha^4=\alpha(\alpha+1)(3\alpha+2)=13\alpha+8\\$

Putting in given expression we get the desired expression equals

$\frac{987\alpha+609}{\alpha^7(\alpha+2)}=\frac{987\alpha+609}{(13\alpha +8)(\alpha+2)}=\frac{987\alpha+609}{(13\alpha^2+34\alpha+16)}=\frac{987\alpha+609}{47\alpha+29}\\$ $=\frac{21(47\alpha+29)}{47\alpha+29}=\boxed{21}$

NOTE: A shortcut method We observe that $\alpha^k=F_{k} \cdot \alpha + F_{k-1}$ where $k$ is natural number and $F_k$ is $k$ th Fibonacci number

So the higher powers of $\alpha$ can be obtained easily by this procedure than the above procedure and then put into the expression.

$\Rightarrow \frac{x^{16} - 1} {x^8 + 2x^7}$

$\Rightarrow \frac{(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)}{x^{7}(x+2)}.....(A)$

Now , we are given ,

$\Rightarrow x^{2}-x-1=0$

We will make some observations ,

$\Rightarrow x^{2}-1=x$

$\Rightarrow (x+1)(x-1)=x$ ..... $(1)$

Also ,

$\Rightarrow x^{2}-1=x$

Squaring it ,

$\Rightarrow (x^{2}-1)^{2}=x^{2}$

$\Rightarrow (x^{4}+1)=3x^{2}$ ..... $(2)$

Also ,

$\Rightarrow (x^{4}+1)^{2}=(3x^{2})^{2}$

$\Rightarrow (x^{8}+1)=7x^{4}$ ..... $(3)$

Substituting $(1),(2)$ and $(3)$ in $(A)$ ,

$\Rightarrow \frac{7x^{4}×3x^{2}×(x^{2}+1)×x}{x^{7}(x+2)}$

$\Rightarrow \frac{7×3×(x^{2}+1)}{(x+2)}.....(B)$

Now again an observation by the given equation ,

$\Rightarrow x^{2}-x-1=0$

$\Rightarrow x^{2}-1=x$

Adding $2$ both sides ,

$\Rightarrow x^{2}+1=x+2$ ..... $(4)$

Placing $(4)$ in $(B)$ we will get ,

$\Rightarrow 7×3×1=\boxed{21}$

This is not a solution.