Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution. 

Let $a_n = x^n + \frac{1}{x^n}$. Then $x^2 = 3x - 1$ implies $a_{n+2} = 3a_{n+1} -a_n$ for all $n$ and also $a_1=3$. Since $a_0=2$, we get
$a_2 = 3a_1 -a_0 = 7$
$a_3 = 3a_2 -a_1 = 18$
$a_4 = 3a_3 -a_2 = 47$
