College Algebra (10th Edition)

$\displaystyle \{-3,\frac{3}{2}-\sqrt{6}i,\frac{3}{2}+\sqrt{6}i\}$
Recognize a sum of cubes, $x^{3}+3^{3}=(x+3)(x^{2}-3x+3^{2})$ One solution is $x=-3$ For the other two, solve $x^{2}-3x+9=0$ Use the quadratic formula $a=1, b=-3, c=9$ $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x=\displaystyle \frac{3\pm\sqrt{9-36}}{2}=\frac{3\pm\sqrt{-24}}{2}$ ...If $N$ is a positive real number, we define the principal square root of $-N,$ denoted by $\sqrt{-N},$ as $\sqrt{-N}=\sqrt{N}i$ $=\displaystyle \frac{3\pm i\sqrt{24}}{2}$ $=\displaystyle \frac{3}{2}\pm\frac{2\sqrt{6}}{2}i$ $=\displaystyle \frac{3}{2}\pm\sqrt{6}i$ Solution set = $\displaystyle \{-3,\frac{3}{2}-\sqrt{6}i,\frac{3}{2}+\sqrt{6}i\}$