<s>[INST] 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. 
 [/INST] Here is an algebraic approach.
$x^2 - 3x + 1 = 0$ implies $\dfrac1x = 3-x$ and so $x^4 + \dfrac{1}{x^4}=x^ 4+(3-x)^4$.
The extended Euclidean algorithm for $\gcd(x^ 4+(3-x)^4,x^2 - 3x + 1)$ gives
$$
47 = 1\cdot(x^ 4+(3-x)^4)+ (-2 x^2 + 6 x - 34)\cdot(x^2 - 3x + 1)
$$
and so $x^4 + \dfrac{1}{x^4}=47$.
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