What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$? 
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$

I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.

Another approach is to note the following
$$\begin{align} x^2&=1-x\\
x^4&=(1-x)^2=1-2x+1-x&=2-3x\\
x^6&=(2-3x)(1-x)=2-5x+3(1-x)&=5-8x\\
x^8&=(5-8x)(1-x)=5-13x+8(1-x)&=13-21x\\
x^{10}&=(13-21x)(1-x)=13-34x+21(1-x)&=34-55x
\end{align}$$
Our fraction becomes
$${49-77x\over 42-66x}={7\over 6}$$
