1. **State the problem:** Find all the cube roots of $-125$. 2. **Recall the formula:** The cube roots of a number $z$ can be found using the polar form and De Moivre's theorem. For a complex number $z = r(\cos \theta + i \sin \theta)$, the $n$th roots are given by: $$ \sqrt[n]{z} = r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right), \quad k=0,1,\ldots,n-1 $$ 3. **Convert $-125$ to polar form:** - Magnitude: $r = |-125| = 125$ - Argument: $\theta = \pi$ (since $-125$ lies on the negative real axis) 4. **Calculate cube roots ($n=3$):** - Magnitude of roots: $125^{1/3} = 5$ - Arguments of roots: $$ \frac{\pi + 2k\pi}{3}, \quad k=0,1,2 $$ 5. **Find each root:** - For $k=0$: $$ 5 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 5 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{5}{2} + i \frac{5\sqrt{3}}{2} $$ - For $k=1$: $$ 5 \left( \cos \pi + i \sin \pi \right) = 5(-1 + 0i) = -5 $$ - For $k=2$: $$ 5 \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = 5 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = \frac{5}{2} - i \frac{5\sqrt{3}}{2} $$ 6. **Final answer:** The three cube roots of $-125$ are: $$ -5, \quad \frac{5}{2} + i \frac{5\sqrt{3}}{2}, \quad \frac{5}{2} - i \frac{5\sqrt{3}}{2} $$