Answer
The three cube roots are
$\,\,\frac{-3}{2}+i\frac{3\sqrt{3}}{2}\,\,\text{ and }\,\,\frac{-3}{2}-i\frac{3\sqrt{3}}{2}$.
Work Step by Step
We will rewrite the provided number as a complex number:
$ z=27+0i $
Now we will convert this into polar form.
For a complex number $ x+iy $, the polar form is $ r\left( \cos \theta +i\sin \theta \right)$. Here,
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& \theta ={{\tan }^{-1}}\left( \frac{y}{x} \right) \\
\end{align}$
For the provided complex number z:
$\begin{align}
& r=\sqrt{{{27}^{2}}+{{0}^{2}}} \\
& =\sqrt{729} \\
& =27
\end{align}$
And,
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{0}{27} \right) \\
& ={{\tan }^{-1}}\left( 0 \right) \\
& =0
\end{align}$
This gives the polar form as:
$ r\left( \cos \theta +i\sin \theta \right)=27\left( \cos 0+i\sin 0 \right)$
Now use deMoivre’s theorem for find the roots,
${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right]$ …… (1)
Now we will put the values of n,r, k and $\theta $ in equation (1) to find out the roots.
For cube roots $ n=3$; k is $0,1,$ and $2$ for three roots.
Now for $ k=0$,
$\begin{align}
& {{z}_{0}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 0}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 0}{3} \right) \right] \\
& ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( 0 \right)+i\sin \left( 0 \right) \right] \\
& =3\left[ 1 \right] \\
& =3
\end{align}$
So the first root of the equation is 3.
For $ k=1$,
$\begin{align}
& {{z}_{1}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 1}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 1}{3} \right) \right] \\
& ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( \frac{2\pi }{3} \right)+i\sin \left( \frac{2\pi }{3} \right) \right] \\
& =3\left[ \frac{-1}{2}+i\frac{\sqrt{3}}{2} \right] \\
& =\frac{-3}{2}+i\frac{3\sqrt{3}}{2}
\end{align}$
The second root of the equation is $\frac{-3}{2}+i\frac{3\sqrt{3}}{2}$.
For $ k=2$,
$\begin{align}
& {{z}_{1}}=\sqrt[3]{27}\left[ \cos \left( \frac{0+2\pi \cdot 2}{3} \right)+i\sin \left( \frac{0+2\pi \cdot 2}{3} \right) \right] \\
& ={{\left( 27 \right)}^{\frac{1}{3}}}\left[ \cos \left( \frac{4\pi }{3} \right)+i\sin \left( \frac{4\pi }{3} \right) \right] \\
& =3\left[ \frac{-1}{2}-i\frac{\sqrt{3}}{2} \right] \\
& =\frac{-3}{2}-i\frac{3\sqrt{3}}{2}
\end{align}$
The third root of the equation is $\frac{-3}{2}-i\frac{3\sqrt{3}}{2}$.
