Question

Find all the 5 roots of the equation x^5 + 243 = 0 .
[ Give your answers in the form of r(cosθ + isinθ). ]

Answer 1

Question

$$\begin{array}{l} x^{5} + 243 = 0 \\ (x + 3)\left(x^{4} - 3x^{3} + 9x^{2} - 27x + 81\right) = 0 \Rightarrow \\ \Rightarrow \\ x = -3 \\ x = 3 \cdot \left(\cos \left(\frac{\pi}{5}\right) + i \cdot \sin \left(\frac{\pi}{5}\right)\right) \\ x = 3 \cdot \left(\cos \left(- \frac{3\pi}{5}\right) + i \cdot \sin \left(- \frac{3\pi}{5}\right)\right) \\ x = 3 \cdot \left(\cos \left(\frac{3\pi}{5}\right) + i \cdot \sin \left(\frac{3\pi}{5}\right)\right) \\ x = 3 \cdot \left(\cos \left(- \frac{\pi}{5}\right) + i \cdot \sin \left(- \frac{\pi}{5}\right)\right) \\ \end{array}$$

Answer: $x = -3$, $x = 3 \cdot \cos \left(\frac{\pi}{5}\right) + 3 \cdot i \cdot \sin$, $x = 3 \cdot \cos \left(-\frac{3\pi}{5}\right) + 3 \cdot i \cdot \sin \left(-\frac{3\pi}{5}\right)$

$$x = 3 \cdot \cos \left(\frac{3\pi}{5}\right) + 3 \cdot i \cdot \sin \left(\frac{3\pi}{5}\right) \text{ and } x = 3 \cdot \cos \left(- \frac{\pi}{5}\right) + 3 \cdot i \cdot \sin \left(- \frac{\pi}{5}\right).$$