Answer
$\left\{2, \dfrac{-1-\sqrt3i}{2}, \dfrac{-1+\sqrt3i}{2}\right\}$
Work Step by Step
The given equation is equivalent to $x^3-2^3=0$.
Factor using the formula for difference of two cubes to obtain:
\begin{align*}
x^3-2^3&=0 \\
(x-2)(x^2+x+1)&=0
\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x-2&=0 &\text{or}& &x^2+x+1=0\\\\
x&=2 &\text{or}& &x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}\\\\
x&=2&\text{or}& &x=\frac{-1\pm\sqrt{1-4}}{2}\\\\
x&=2&\text{or}& &x=\frac{-1\pm\sqrt{-3}}{2}\\\\
x&=2&\text{or}& &x=\frac{-1\pm \sqrt{3}i}{2}\\\\
\end{align*}
Thus, the solutions are:
$\left\{2, \dfrac{-1\pm\sqrt3i}{2}\right\}$
