One Two Three

$\begin{aligned} x^3 + \frac 1{x^3} & = \left(x + \frac 1x \right) \left(\color{#3D99F6}{x^2 + \frac 1{x^2}} - 1 \right) & \small \color{#3D99F6}{\text{Given that } x + \frac 1x = x^2 + \frac 1{x^2}} \\ & = \left(x + \frac 1x \right) \left(\color{#3D99F6}{x + \frac 1x} - 1 \right) \\ & = x^2 + 1 - x + 1 + \frac 1{x^2} - \frac 1x \\ & = \cancel{x^2 + \frac 1{x^2}} - \cancel{\left(x + \frac 1x\right)} + 2 \\ & = \boxed{2} \end{aligned}$

Let

$x + \frac{1}{x} = a$

So

${(x + \frac{1}{x})}^{2} = a^2$

$x^2 + \frac{1}{x^2} = a^2 - 2$

From the question

$(a^2 -2 = a) > 0$

$a^2 -a -2 = 0$

$(a-2)(a+ 1) =0$

$\text{We know a >0} \therefore a = 2$

$x + \frac{1}{x} = 2$

So

$(x-1)^2 = 0 \implies x = 1$

$\therefore x^3 + \dfrac1{x^3} = 1$