<s>[INST] What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$. 
How to find roots here?
 [/INST] Put $x = y - 1$, this yields:
$$y^3 + y - 1 = 0$$
Then compare with the identity:
$$\begin{split}
(a+b)^3 \equiv a^3 + 3 a^2 b + 3 b a^2 + b^3\Longrightarrow \\
(a+b)^3 - 3 a b (a+b) - (a^3 + b^3) \equiv 0
\end{split}
 $$
So, if we can find two numbers $a$ and $b$ such that:
$$-3 a b = 1$$
and
$$a^3 + b^3 = 1$$
then $y = a + b$ will be a solution.
If we put $A = a^3$ and $B = b^3$, then we require that:
$$A B = -\frac{1}{27}$$
$$A + B = 1$$
Now solving for $A$ and $B$ requires solving a quadratic equation. When extracting $a$ and $b$, you must make sure that $a b = -\frac{1}{3}$, so the choice of the phase factor multiplying the cube root for $A$ fixes the phase factor for $b$.
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