<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] 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 $\zeta_n=e^{\frac{2\pi ik}{n}}$
So the roots are
$$x_1=(2)^\frac{1}{3} e^\frac{2\pi .1}{3}=(2)^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1),$$
$$x_3=(2)^\frac{1}{3} e^\frac{2\pi .2}{3}=(2)^\frac{1}{3},$$
$$x_2=(2)^\frac{1}{3} e^\frac{2\pi .2}{3}=
(2)^\frac{1}{3}(-\frac{1}{2})(\sqrt{3}i+1).$$
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