WMC 2018 Senior Problem 23

Given that $x = 3 + \sqrt 8 = \dfrac {6 + \sqrt{6^2-4(1)}}2$ , implying $3+\sqrt 8$ is a root of $x^2 - 6x + 1=0$ . Therefore,

$\begin{aligned} x^2 - 6x + 1 & = 0 & \small \color{#3D99F6} \text{Divide both sides by }x. \\ x - 6 + \frac 1x & = 0 \\ x + \frac 1x & = 6 & \small \color{#3D99F6} \text{Squaring both sides.} \\ x^2 + 2 + \frac 1{x^2} & = 36 \\ x^2 + \frac 1{x^2} & = 34 & \small \color{#3D99F6} \text{Similarly, squaring and }- 2\text{ both sides.} \\ \implies x^4 + \frac 1{x^4} & = 34^2 - 2 \\ & = \boxed{1154} \end{aligned}$

Given that $x = 3 + \sqrt 8$ . So $\dfrac 1x = \dfrac 1{3 + \sqrt 8} = \dfrac{3 - \sqrt 8}{(3 + \sqrt 8)(3 - \sqrt 8)} = \dfrac{(3 - \sqrt 8)}{3^2 - (\sqrt 8)^2} = 3 - \sqrt 8$

Now,

$\begin{aligned} x + \dfrac 1x & = 3+ \sqrt 8 + 3 - \sqrt 8 \\ \Rightarrow \left(x + \dfrac 1x \right)^2 & = 6^2 ~~~~~[\text{Square on both sides}] \\ \Rightarrow x^2 + \dfrac 1{x^2} +2 & = 6^2 \\ \Rightarrow x^2 + \dfrac 1{x^2} & = 34 \\ \Rightarrow \left(x^2 + \dfrac 1{x^2} \right)^2 & = 34^2 ~~~~~~~ [ \text{Square on both sides}] \\ \Rightarrow (x^2)^2 + \dfrac 1{(x^2)^2} + 2 & = 34^2 \\ \Rightarrow x^4 + \dfrac 1{x^4} & = 1156 - 2 \\ \implies x^4 + \dfrac 1{x^4} & = \boxed{1154} \\ \end{aligned}$