Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$ 
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$

The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?

A polynomial $f(x)$ is a multiple of $x^2-x-1$ (the characteristic polynomial of the Fibonacci sequence) iff $f(\varphi)=f(\bar{\varphi})=0$, with $\varphi=\frac{1+\sqrt{5}}{2}$, $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$ being algebraic conjugates. Since $\varphi^2=\varphi+1$ we have by induction $\varphi^k=F_{k}\varphi+F_{k-1}$ and the same holds for $\bar{\varphi}$. It follows that $ax^9+bx^8+1$ is a multiple of $x^2-x-1$ iff
$$\begin{eqnarray*} 0 &=& a\varphi^9+b\varphi^8+1 = a(34\varphi+21)+b(21\varphi+13)+1\\&=&(34a+21b)\varphi+(21a+13b+1)\end{eqnarray*}$$
and
$$ 0 = (34a+21b)\bar{\varphi}+(21a+13b+1). $$
It follows that we must have $21a+13b=-1$ and $34a+21b=0$, so $\color{red}{a=21,b=-34}$ works.
You may also invoke the more general identity
$$ (x^2-x-1)\sum_{k=0}^{n}(-1)^{k+1}F_k x^k = (-1)^{n+1}F_n x^{n+2}+(-1)^n F_{n+1} x^{n+1}-x.$$
