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. 

Clearly, $x\ne0$ in $x^2-3x+1=0$
Dividing both sides by $x,$
$$\dfrac{x^2-3x+1}x=0\implies x+\dfrac1x=3$$
$$x^{2n}+\dfrac1{x^{2n}}=\left(x^n+\dfrac1{x^n}\right)^2-2$$
