<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] Let's say $a > 0$ is a real number.
The roots of $x^3 = a$ are these
$x_1 = a^\frac{1}{3} ( \cos{2\pi/3} + i \cdot \sin{2\pi/3} )$
$x_2 = a^\frac{1}{3} ( \cos{4\pi/3} + i \cdot \sin{4\pi/3} )$
$x_3 = a^\frac{1}{3} ( \cos{6\pi/3} + i \cdot \sin{6\pi/3} )$
We can simplify these formulas further...
The root $x_3$ happens to be real.
The other two are complex conjugated to each other.
Seems your problem is that your knowledge of the roots of unity is a bit rusty.
You can refresh it here:
Roots of unity
Once you know how to find the roots of $x^3=1$,
what's left to do is just to add a factor of $a^{\frac{1}{3}}$.
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