Number

$\begin{aligned} x^2 +x + 1 & = 0 \\ \Rightarrow x^2 + 1 & = - x \\ \Rightarrow \dfrac{x^2 + 1 }x & = - 1 \\ \Rightarrow x + \dfrac 1x & = -1 \\ \Rightarrow \left(x + \dfrac 1x \right)^3 & = (-1)^3 ~~~~~ [\text{Cube both sides}] \\ \Rightarrow x^3 + \dfrac 1{x^3} + 3\left(x + \dfrac 1x \right) & = - 1 \\ \Rightarrow x^3 + \dfrac 1{x^3} - 3 & = - 1 ~~~~~~~~~ [ x + \dfrac 1x = - 1] \\ \implies x^3 + \dfrac 1{x^3} & = \boxed 2 \\ \end{aligned}$

The roots of the above quadratic equation equal:

$x = \frac{-1 \pm \sqrt{3}i}{2} = e^{\pm i \frac{2\pi}{3}}$

which if we substitute into the sum of cubes, we obtain:

$x^3 + \frac{1}{x^3} = (e^{i\frac{2\pi}{3}})^3 + (e^{-i\frac{2\pi}{3}})^3 = e^{i 2\pi} + e^{-i 2\pi} = 2 \cdot cos(2\pi) = \boxed{2}.$

Because x^2 + x + 1 = 0,

x^3 + x^2 + x = 0.

Now subtract from the 2 equations:

x^3 - 1 = 0

So x^3 = 1.

Now, x^3 + 1/x^3 = 1 + 1 = 2.