+118667 what are the zeros of this equation? x^3+1000
\(x^3+1000=0\\ x^3=-1000\\ x=-10 \)
x=-10 is the only real solution but since it is a cubic there will be two more imaginary roots.
they well be
\(x=cos(\pi-\frac{2\pi}{3})+isin(\pi-\frac{2\pi}{3})\quad and \quad x=cos(\pi+\frac{2\pi}{3})+isin((\pi+\frac{2\pi}{3})\\ x=cos(\frac{\pi}{3})+isin(\frac{\pi}{3})\quad and \quad x=cos(\frac{5\pi}{3})+isin(\frac{5\pi}{3})\\ x=\frac{1}{2}+\frac{\sqrt 3}{2}\;i\quad and \quad x=\frac{1}{2}-\frac{\sqrt 3}{2}\;i\\ x=\frac{(1+\sqrt 3)i}{2}\;\quad \;\;and \quad x=\frac{(1-\sqrt 3)i}{2}\\ \)
