Value matters#2

$\begin{aligned} \left(x+\frac 1x\right)^2 & = x^2 + 2 + \frac 1{x^2} = 2+3 = 5 \\ \implies x + \frac 1x & = \pm \sqrt 5 \end{aligned}$

Now, we have:

$\begin{aligned} \left(x+\frac 1x\right)^3 & = x^3 + 3x + \frac 3x + \frac 1{x^3} \\ x^3 + \frac 1{x^3} & = \left({\color{#3D99F6} x+\frac 1x} \right)^3 - 3 \left({\color{#3D99F6} x+\frac 1x} \right) & \small \color{#3D99F6} \text{Using the positive value }x+\frac 1x = \sqrt 5 \\ \implies \frac {x^6+1}{x^3} & = \left({\color{#3D99F6} \sqrt 5} \right)^3 - 3 \left({\color{#3D99F6} \sqrt 5} \right) = \boxed{2\sqrt{5}} \end{aligned}$