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.

$$x^2=1-x$$
$$x^4=(1-x)^2=x^2-2x+1=1-x-2x+1=2-3x$$
$$x^6=(1-x)(2-3x)=3x^2-5x+2=3(1-x)-5x+2=5-8x$$
$$x^8=(2-3x)^2=9x^2-12x+4=9(1-x)-12x+4=13-21x$$
$$x^{10}=(1-x)(13-21x)=21x^2-34x+13=21(1-x)-34x+13=34-55x$$
Therefore,
$$x^{10}+x^8+x^2+1=34-55x+13-21x+1-x+1=49-77x$$
$$x^{10}+x^6+x^4+1=34-55x+5-8x+2-3x+1=42-66x$$
Your equation 
=$$\frac{49-77x}{42-66x}=\frac{7}{6}$$
