You may use algebra if you want

$x^2 - x - 1 = 0 \Rightarrow x^2 - 1 = x \Rightarrow x - \frac 1x = 1 \\ \Rightarrow \left (x - \frac 1x \right )^2 = 1 \Rightarrow x^2 + \frac 1{x^2} = 3 \\ \Rightarrow \left ( x^2 + \frac 1{x^2} \right )^2 = 3^2 \Rightarrow x^4 + \frac 1 {x^4} = 7 \\ \Rightarrow x^4 + 1 = 3x^2, x^8 + 1 = 7x^4$

$\begin{aligned} \frac {x^{16} - 1}{x^8 + 2x^7} &=& \frac {(x^2 - 1)(x^2 + 1)(x^4 + 1)(x^8 + 1) }{x^7(x+2)} \\ &=& \frac {(x)(x+2)(3x^2)(7x^4) }{x^7(x+2)} \\ &=& \boxed{21} \\ \end{aligned}$

A root of $x^2 - x - 1 = 0$ is the golden ratio, $\varphi = \frac {1+\sqrt{5}}{2}$ and its powers satisfy the following equation.

$\varphi^n = F_n\varphi + F_{n-1}$ for $n=1,2,3,...$ , where $F_n$ is the $n^{th}$ Fibonacci number and $F_0 = 0$ .

With an Excel spreadsheet we can easily find $\varphi^n$ :

\(\begin{array} {} \varphi \space \space = \varphi & \varphi^{5} = 5 \varphi + 3 & \varphi^{9} = 34 \varphi + 21 & \varphi^{13} = 233 \varphi + 144 \\ \varphi^{2} = \varphi + 1 & \varphi^{6} = 8 \varphi + 5 & \varphi^{10} = 55 \varphi + 34 & \varphi^{14} = 377 \varphi + 233 \\ \varphi^{3} = 2 \varphi + 1 & \varphi^{7} = 13 \varphi + 8 & \varphi^{11} = 89 \varphi + 55 & \varphi^{15} = 610 \varphi + 377 \\ \varphi^{4} = 3 \varphi + 2 & \varphi^{8} = 21 \varphi + 13 & \varphi^{12} = 144 \varphi + 89 & \varphi^{16} =987 \varphi + 610 \end{array} \)

$\Rightarrow \dfrac {x^{16}-1}{x^8+2x^7} = \dfrac {987 \varphi + 610 - 1}{21 \varphi + 13+2(13 \varphi + 8)} = \dfrac {987\varphi+609}{47\varphi+29} \\ \quad = \dfrac {21(47\varphi+29)}{47\varphi+29} = \boxed{21}$

From

$x^2 - x - 1 = 0 \implies x^2 = x + 1$

Then,

$\begin{aligned} x^{16} - 1 &= (x^{8}+ 1) \cdot (x^{4} + 1) \cdot (x^{2} + 1 ) \cdot (x^2 - 1) \\ &= (x^{8}+ 1) \cdot (x^{4} + 1) \cdot (x+2) \cdot (x) \end{aligned}$

and

$\begin{aligned} x^8 + 2x^7 &= (x^6) \cdot (x^2 + 2x) \\ &= (x^6) \cdot (x + 2) \cdot (x) \end{aligned}$

Hence,

$\begin{aligned} \dfrac{x^{16} - 1}{x^8 + 2x^7} &= \dfrac{ (x^{8}+ 1) \cdot (x^{4} + 1) \cdot (x+2) \cdot (x) }{ (x^6) \cdot (x + 2) \cdot (x) } \\ &= \dfrac{ x^{12} + x^8 + x^4 + 1 }{ x^6 } \\ &= x^6 + \dfrac{1}{x^6} + x^2 + \dfrac{1}{x^2} \end{aligned}$

From Mr. @Pi Han Goh 's Solution,

$x^2 + \dfrac{1}{x^2} = 3$

Then,

$\begin{aligned} x^6 + \dfrac{1}{x^6} &= \left( x^2 + \dfrac{1}{x^2} \right) \left(x^4 - 1 + \dfrac{1}{x^4} \right) \\ &= \left( x^2 + \dfrac{1}{x^2} \right) \left(\left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 - 1 \right) \\ &= (3)(3^2 - 3) = 18 \end{aligned}$

Therefore,

$\begin{aligned} \dfrac{x^{16} - 1}{x^8 + 2x^7} &= x^6 + \dfrac{1}{x^6} + x^2 + \dfrac{1}{x^2} \\ &= 3 + 18 = \boxed{21} \end{aligned}$