<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] If we divide both sides by $x$ we get $$\frac{x^2-3x+1}{x} = x-3+\frac1x = 0\implies x+\frac1x = 3$$Squaring both sides$$\left(x+\frac1x\right)^2 = x^2+2+\frac1{x^2} = 9\implies x^2+\frac1{x^2}=7$$Squaring again$$\left(x+\frac1x\right)^4 = \left(x^2+\frac1{x^2}\right)^2 = x^4+2+\frac1{x^4} = 49\implies x^4+\frac{1}{x^4}=\boxed{47}$$
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