# Question 0c521

Apr 8, 2017

Let

${x}^{3} = 64 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right)$

$\implies {x}^{3} = 64 \left(- 1 + i \times 0\right)$

$\implies {x}^{3} + 64 = 0$

$\implies {x}^{3} + {4}^{3} = 0$

$\implies \left(x + 4\right) \left({x}^{2} - 4 x + 16\right) = 0$

So when $x + 4 = 0$

$\implies x = - 4 = 4 \left(- 1 + i \times 0\right) = 4 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right)$

when

$\left({x}^{2} - 4 x + 16\right) = 0$

$\implies x = \frac{4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot 16}}{2}$

$\implies x = \frac{4 \pm \sqrt{- 48}}{2}$

$\implies x = \frac{4 \pm 4 \sqrt{- 3}}{2}$

$\implies x = \left(2 \pm i 2 \sqrt{3}\right)$

$\implies x = 4 \left(\frac{1}{2} \pm i \frac{\sqrt{3}}{2}\right)$

So $\implies x = 4 \left(\cos \left(\frac{\pi}{3}\right) \pm i \sin \left(\frac{\pi}{3}\right)\right)$

So three cube roots are

$4 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right) , 4 \left(\cos \left(\frac{\pi}{3}\right) \pm i \sin \left(\frac{\pi}{3}\right)\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Applying de moivre's formula

we have

${x}^{3} = 64 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right)$

$x = {64}^{\frac{1}{3}} \left(\cos \left(\frac{\pi + 2 \pi k}{3}\right) + i \sin \left(\frac{\pi + 2 \pi k}{3}\right)\right)$

where k varies over the integer values from 0 to 2

For $k = 0$

$x = {64}^{\frac{1}{3}} \left(\cos \left(\frac{\pi + 2 \pi \times 0}{3}\right) + i \sin \left(\frac{\pi + 2 \pi \times 0}{3}\right)\right)$

$\implies x = 4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$

For $k = 1$

$\implies x = {64}^{\frac{1}{3}} \left(\cos \left(\frac{\pi + 2 \pi \times 1}{3}\right) + i \sin \left(\frac{\pi + 2 \pi \times 1}{3}\right)\right)$

=>x= 4(cos((pi)+isin((pi))#

For $k = 2$

$x = {64}^{\frac{1}{3}} \left(\cos \left(\frac{\pi + 2 \pi \times 2}{3}\right) + i \sin \left(\frac{\pi + 2 \pi \times 2}{3}\right)\right)$

$\implies x = 4 \left(\cos \left(2 \pi - \frac{\pi}{3}\right) + i \sin \left(2 \pi - \frac{\pi}{3}\right)\right)$

$\implies x = 4 \left(\cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right)$