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. 
 A: 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$$\left(x+\frac1x\right)^4 = \left(x^2+\frac1{x^2}\right)^2 = x^4+2+\frac1{x^4} = 49\implies x^4+\frac{1}{x^4}=\boxed{47}$$
A: Clearly, $x\ne0$ in $x^2-3x+1=0$
Dividing both sides by $x,$
$$\dfrac{x^2-3x+1}x=0\implies x+\dfrac1x=3$$
$$x^{2n}+\dfrac1{x^{2n}}=\left(x^n+\dfrac1{x^n}\right)^2-2$$
A: Hint Dividing gives $$x - 3 + \frac{1}{x} = 0,$$ and then rearranging gives
$$x + \frac{1}{x} = 3.$$

Additional hint Squaring gives $$3^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2},$$ and rearranging again gives $$x^2 + \frac{1}{x^2} = 7.$$

A: You can just use 'brute force'. Using your favorite method of solving quadratics, the second equation gives you:
$$
x=\frac{3\pm \sqrt{5}}{2}
$$
Then use this value to calculate $x^4+\frac{1}{x^4}=47$.
A: 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
$a_2 = 3a_1 -a_0 = 7$
$a_3 = 3a_2 -a_1 = 18$
$a_4 = 3a_3 -a_2 = 47$
A: 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 $x^4 + \dfrac{1}{x^4}=47$.
