<s>[INST] 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!
 [/INST] The best way to do it is to use the methods that other responses have detailed.  I am going to show you an inferior method only because understanding the relationship between this inferior method and the better method (already described by others) will stretch your intuition.
Let $a$ denote $2^{(1/3)}.$
Given $0 = (x^3 - 2) = (x^3 - a^3) = (x - a)(x^2 + ax + a^2)$, then
one of the roots will be given by the factor $(x - a)$.
The other two roots will come from applying the quadratic eqn against $(x^2 + ax + a^2) = 0$.
This gives $$x = \frac{1}{2} \times \left[-a \pm\sqrt{a^2 - 4a^2}\right]$$
$$= \frac{1}{2} \times \left[-a \pm i \times \sqrt{3} \times a\right]$$
$$= \frac{1}{2} \times a \times \left[-1 \pm i \times \sqrt{3}\right]$$
$$= 2^{(1/3)} \times \frac{1}{2} \times \left[-1 \pm i \times \sqrt{3}\right].$$
</s>