## College Algebra (10th Edition)

$\{2,-1-\sqrt{3}i,-1+\sqrt{3}i\}$
Recognize a difference of cubes, $x^{3}-2^{3}=(x-2)(x^{2}+2x+2^{2})$ One solution is $x=2$. For the other two, solve $x^{2}+2x+4=0$ Use the quadratic formula $a=1, b=2, c=4$ $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x=\displaystyle \frac{-2\pm\sqrt{4-16}}{2}=\frac{-2\pm\sqrt{-12}}{2}$ ...If $N$ is a positive real number, we define the principal square root of $-N,$ denoted by $\sqrt{-N},$ as $\sqrt{-N}=\sqrt{N}i$ $=\displaystyle \frac{-2\pm i\sqrt{12}}{2}$ $=-\displaystyle \frac{2}{2}\pm\frac{2\sqrt{3}}{2}i$ $=-1\pm\sqrt{3}i$ Solution set = $\{2,-1-\sqrt{3}i,-1+\sqrt{3}i\}$