Not Defined

As it is clear that $x \ne 0$ , we can divide the given equation through by $x$ to find that

$x - 1 + \dfrac{1}{x} = 0 \Longrightarrow x + \dfrac{1}{x} = 1$ .

We then have that $\left(x + \dfrac{1}{x}\right)^{2} = 1 \Longrightarrow x^{2} + \dfrac{1}{x^{2}} + 2 = 1 \Longrightarrow x^{2} + \dfrac{1}{x^{2}} = -1$ .

So finally $x^{3} + \dfrac{1}{x^{3}} = \left(x + \dfrac{1}{x}\right)\left(x^{2} + \dfrac{1}{x^{2}}\right) - \left(x + \dfrac{1}{x}\right) = (1 \times (-1)) - 1 = \boxed{-2}$ .

$x^2 - x + 1 = 0$

$\Longrightarrow x^2 = x-1$

$\Longrightarrow x^3 = x^2-x$

$\Longrightarrow x^3 = (x-1)-x$

$\Longrightarrow x^3 = -1$

$\Longrightarrow x^3 + \frac{1}{x^3} = -2$

The roots of the given equation are: $x=\frac{1 \pm i \sqrt3} {2} = e^{\frac {\pm\pi i} {3}}$ . If you raise that to the power of 3, you get: $(e^{\frac {\pm\pi i} {3}}) ^3 = e^{\pi} = - 1$ . Inserting this in the question at hand gives: $-1 + \frac{1} {-1} =-2$ .

$x^2 -x +1 = 0 \Rightarrow (x+1)(x^2 -x +1) = 0(x+1)$ . So, $x^3 +1= 0 \Rightarrow x^3 = -1$ . So, $x^3 + \frac{1}{x^3} = -1+ \frac{1}{-1} = -2$ .

$x^2-x+1=0$

$\Rightarrow \dfrac{x^3+1}{x+1}=0$

as $x\neq-1$ we have,

$x^3+1=0$ or $x^3=-1$

$x^3+\dfrac{1}{x^3}=\boxed{-2}$

$\begin{aligned} x^2 - x + 1 & = 0 & \small \color{#3D99F6} \text{Divide both sides by }x \\ x - 1 + \frac 1x & = 0 & \small \color{#3D99F6} \text{Rearrange} \\ x + \frac 1x & = 1 & \small \color{#3D99F6} \text{Cube both sides} \\ \left(x+\frac 1x \right)^3 & = 1^3 \\ x^3 + 3x + \frac 3x + \frac 1{x^3} & = 1 & \small \color{#3D99F6} \text{Rearrange} \\ \implies x^3 + \frac 1{x^3} & = 1 - 3\left({\color{#3D99F6}x + \frac 1x} \right) & \small \color{#3D99F6} \text{Note that }x + \frac 1x = 1 \\ & = 1 - 3({\color{#3D99F6}1}) \\ & = \boxed{-2} \end{aligned}$