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?

Consider
$$\frac{a x^9+bx^8+1}{x^2-x-1}$$ perform long division to get
$$-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7-(b+34) x^8+x^9
   (-a+b+55)+O\left(x^{10}\right)$$
So $b=-34$ and $a=21$ and the result is
$$\frac{a x^9+bx^8+1}{x^2-x-1}=-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7$$
