TITLE: What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: $$\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} =\dfrac{(x^2+1)(x^8+1)}{(x^4+1)(x^6+1)}$$
$$=\dfrac{\left(x+\dfrac1x\right)\left(x^4+\dfrac1{x^4}\right)}{\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right)}$$
$$=\dfrac{\left(x^2+\dfrac1{x^2}\right)^2-2}{\left(x^2+\dfrac1{x^2}\right)\left(x^2+\dfrac1{x^2}+1\right)}$$
Now $x^2+x-1=0\implies x-\dfrac1x=-1$ as $x\ne0$
Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=1+2$
Can you take it from here?