<s>[INST] Quick way to find all roots of a polynomial? I'm following an example of finding the splitting field for $f(X)=X^3-2$ over $\mathbb{Q}$. It reads, if $\alpha$ is the positive cube root of $2$, then the roots of $f$ are $\alpha$, $\alpha(-\frac{1}{2}+i\frac{1}{2}\sqrt{3})$, and $\alpha(-\frac{1}{2}-i\frac{1}{2}\sqrt{3})$.
My question is, how does one know what those two latter roots are so quickly? Is there a general way to find all the other roots of a polynomial if you can find the obvious one, in this case $\sqrt[3]{2}$? 
 [/INST] The other two roots are $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$, where $\omega$ is a primitive cubic root of unity. It just so happens that the primitive cubic roots of unity are easy to express, because they are the "other two" roots of $x^3-1 = (x-1)(x^2+x+1)$, so they can be found using the quadratic formula.
In general, the $n$ complex roots of $x^n-a$, with $a\in\mathbb{R}$, are given by
$$\sqrt[n]{a},\quad \zeta_n\sqrt[n]{a},\quad \zeta_n^2\sqrt[n]{a},\quad\ldots,\quad \zeta_n^{n-1}\sqrt[n]{a},$$
where $\zeta_n$ is a primitive $n$th root of unity.

Also, if you happen to know one root of a cubic polynomial, then you can always divide and solve the resulting quadratic. Here, you have $x^3-2$, and you know that $x-\sqrt[3]{2}$ is a factor (because $\sqrt[3]{2}$ is a root). Factoring, you have
$$x^3 - 2 = (x-\sqrt[3]{2})(x^2 + \sqrt[3]{2}x + \sqrt[3]{4}).$$
So the other two roots are the roots of the quadratic, which can be found using the quadratic formula:
$$r_1 = \frac{-2^{1/3} + \sqrt{2^{2/3} - 4\cdot 2^{2/3}}}{2},\qquad r_2 = \frac{-2^{1/3} - \sqrt{2^{2/3} - 4\cdot 2^{2/3}}}{2}.$$
Simplify the square root, factor out $2^{1/3}$, and you get the expressions as well. E.g.,
$$\begin{align*}
r_1 &= \frac{-2^{1/3}+\sqrt{2^{2/3}-4\cdot 2^{2/3}}}{2}\\
&= \frac{-2^{1/3}+\sqrt{2^{2/3}(-3)}}{2}\\
&= \frac{-2^{1/3} + 2^{1/3}\sqrt{-3}}{2}\\
&= \left(\frac{-1+\sqrt{-3}}{2}\right)2^{1/3}\\
&= \left( - \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\sqrt[3]{2}.
\end{align*}$$
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