<s>[INST] 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.
 [/INST] Applying polynomial long division, we have
$\begin{array}{rlllllllllll}
&~~1x^8-1x^7+3x^6+\dots\\
\hline
x^2+x-1&|x^{10}+0x^9+x^8+0x^7+0x^6+0x^5+0x^4+0x^3+1x^2+0x+1\\
&~x^{10}+x^9-x^8\\
\hline
&~~~~~-x^9+2x^8+0x^7\\
&~~~~~-x^9-x^8+x^7\\
\hline
&~~~~~~~~~~~~~~~~~3x^8+\dots\\
&~~~~~~~~~~~~~~~~~~\vdots
\end{array}$

eventually arriving at $$x^{10}+x^8+x^2+1\\=(x^8-x^7+3x^6-4x^5+7x^4-11x^3+18x^2-29x+48)(x^2+x-1)+49-77x\\=49-77x$$

Similarly, applying polynomial long division to the denominator, it simplifies to 

$42-66x$.

So the fraction in general simplifies to

$\frac{49-77x}{42-66x}=\frac{7}{6}$

</s>