## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\left\{-1, \dfrac{1-\sqrt3i}{2}, \dfrac{1+\sqrt3i}{2}\right\}$
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: $\left\{-1, \dfrac{1\pm\sqrt3i}{2}\right\}$