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

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

A: $$\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} =\dfrac{(x^2+1)(x^8+1)}{(x^4+1)(x^6+1)}$$
$$=\dfrac{\left(x+\dfrac1x\right)\left(x^4+\dfrac1{x^4}\right)}{\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right)}$$
$$=\dfrac{\left(x^2+\dfrac1{x^2}\right)^2-2}{\left(x^2+\dfrac1{x^2}\right)\left(x^2+\dfrac1{x^2}+1\right)}$$
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?
A: HINT: Note that $$\color{Green}{x^2=-x+1}.$$
By multiplying the given expression by $x,$ we can obtained $$\color{Green}{x^3=2x-1}.$$ Again by multiplying $x$ we have $$\color{Green}{x^4=-3x+2}$$ and so on..
A: 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
$${49-77x\over 42-66x}={7\over 6}$$
