Answer
$\left\{-1, \dfrac{1-\sqrt3i}{2}, \dfrac{1+\sqrt3i}{2}\right\}$
Work Step by Step
The given equation is equivalent to $x^3+1^3=0$.
Factor using the formula for sum of two cubes to obtain:
\begin{align*}
x^3+1^3&=0 \\
(x+1)(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+1&=0 &\text{or}& &x^2-x+1=0\\\\
x&=-1 &\text{or}& &x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}\\\\
x&=-1&\text{or}& &x=\frac{1\pm\sqrt{1-4}}{2}\\\\
x&=-1&\text{or}& &x=\frac{1\pm\sqrt{-3}}{2}\\\\
x&=-1&\text{or}& &x=\frac{1\pm \sqrt{3}i}{2}\\\\
\end{align*}
Thus, the solution set is:
$\left\{-1, \dfrac{1\pm\sqrt3i}{2}\right\}$
