How do you find the three cube roots of unity?

Hint: We first assume the cube roots of unity as variable x and try to form the cubic equation which gives the roots. Then we solve the factorisation to find the roots out of which one is real root and the other two are imaginary.

Complete step by step solution:
We have to find cube roots of unity where we assume the roots as variable x. Here the variable can be any one out of the three roots.
For any value of the variable, we have the cubic value of the variable as 1.
Therefore, ${{x}^{3}}=1$. The equation becomes ${{x}^{3}}-1=0$.
Using the identity of ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ for ${{x}^{3}}-1=0$, taking $a=x,b=1$ we get
${{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0$.
Multiplication of two terms is 0 which gives either $\left( x-1 \right)=0$ or $\left( {{x}^{2}}+x+1 \right)=0$.
When $\left( x-1 \right)=0$, then we get $x=1$, one real root out of three cube roots of unity.
Now when $\left( {{x}^{2}}+x+1 \right)=0$, we solve it using quadratic solving.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $\left( {{x}^{2}}+x+1 \right)=0$. The values of a, b, c are $1,1,1$ respectively.
We put the values and get $x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 1\times 1}}{2\times 1}=\dfrac{-1\pm \sqrt{-3}}{2}=\dfrac{-1\pm \sqrt{3}i}{2}$.
Here $i$ is the imaginary number in unity.
The other two imaginary roots are $x=\dfrac{-1\pm \sqrt{3}i}{2}$ which are termed as $\omega ,{{\omega }^{2}}$.
Therefore, cube roots of unity are $x=1,\dfrac{-1\pm \sqrt{3}i}{2}$.

Note: There are some properties related to the cube roots of unity where we get $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$. The value of $\omega ,{{\omega }^{2}}$ can be any one of the $\dfrac{-1\pm \sqrt{3}i}{2}$.The root of unity is a number which is complex in nature and gives 1 if raised to the power of a positive integer n.