Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$,  $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi i}{n}}$
So the roots are $2^\frac{1}{3}, -(2^\frac{1}{3}) ~and~ 2^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1).$
Is this correct?
Note: I don't like factoring method. I studied in schools time. So don't help with that method.
Thanks!

Forward
i) Euler's formula: $e^{i\theta}=\cos\theta + i\sin\theta$
ii) De Moivre's formula: $\left(\cos x+i\sin x\right)^n=\cos (nx) + i\sin (nx)$
Theorem
If $z\ne 0$, and if $n$ is a positive integer, then there are exactly $n$ distinct complex numbers $z_0,z_1,\dots,z_{n-1}$ such that $z_k^n=z$, $k=0,\dots,n-1$. These roots are given by the formula:
$$z_k=\lvert{z}\rvert^{\frac1n}\exp\left[ i\left(\frac{\text{arg}(z)}{n}+\frac{2\pi k}{n}\right)\right],\qquad k=0,\dots,n-1$$
(e.g. Apostol, Mathematical Analysis, Theorem 1.51)
Roots of unity
Since $\lvert 1\rvert=1$ and $\text{arg}(1)=0$, the $n$ roots of unity are:
\begin{align*}z_k&=1^{\frac1n}\exp\left[i\left(\frac{\text{arg}(z)}{n}+\frac{2\pi k}{n}\right)\right]=\exp\left(i\frac{2\pi k}{n}\right)\\&=\cos\frac{2\pi k}{n}+i\sin\frac{2\pi k}{n}\end{align*}
When $n=3$, the three roots of unity are:
i) $k=0$: $$z_0=\cos 0+i\sin 0=1$$and $1^3=1$;
ii) $k=1$: $$z_1=\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3}=\frac{-1+i\sqrt{3}}{2}$$and $\left(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)^3=\cos (2\pi)+i\sin(2\pi)=1$;
iii) $k=2$: $$z_2=\cos \frac{4\pi}{3}+i\sin\frac{4\pi}{3}=\frac{-1-i\sqrt{3}}{2}$$and $\left(\cos \frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)^3=\cos (4\pi)+i\sin(4\pi)=1$.
Roots of a real positive number
If $x^3=2$, then the three roots are:
$$x_0=2^\frac13,\quad x_1=2^\frac13\left(\frac{-1+i\sqrt{3}}{2}\right),\quad x_2=2^\frac13\left(\frac{-1-i\sqrt{3}}{2}\right)$$
