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?
 A: This can be done by hand with a bunch of substitutions. 
Let $y=x+1$, then
$$1+4(y-1)+3(y-1)^2+(y-1)^3=0$$
If we expand all the terms we get
$$1+4y-4+3y^2-6y+3+y^3-3y^2+3y-1=0$$
which boils down to 
$$y^3+y-1=0$$
Now we can do a change in coordinates with another substitution. 
Let $y=u+\frac{\lambda}{u}$. We will find $\lambda$ later. This yields. 
$$-1+u+\frac{\lambda}{u}+\left ( u + \frac{\lambda}{u} \right )^3=0$$
We can expand this by multiplying by $u^3$ which gives
$$u^4(3\lambda + 1) + u^2\lambda(3\lambda+1)-u^3+u^6+\lambda^3=0$$
We want the coefficients on the $u^2$ and $u^4$ terms to be $0$. So $\lambda=-\frac{1}{3}$. 
Substituting we now have
$$u^6-u^3-\frac{1}{27}=0$$
If we perform yet another substitution, we can finally get to a quadratic.
Let $z=u^3$. 
$$z^2-z+\frac{1}{27}=0$$ 
Using the quadratic formula we get
$$z=\frac{1}{18}(9+\sqrt{93})$$.
Now we climb back up our ladder of subsitutions
$$u^3=\frac{1}{18}(9+\sqrt{93})$$
$$u=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}$$
We substitute for $u$ and get
$$y=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}-\frac{1}{3\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}}$$
Now to get $x$ we just subtract 1. 
$$x=\left( \frac{1}{18}(9+\sqrt{93})\right )^{1/3} - \frac{1}{3}\left( \frac{1}{18}(9+\sqrt{93})\right )^{-1/3}-1$$
A: Maybe this could help a bit?
http://www.wolframalpha.com/input/?i=Root[x^3%2B3*x^2%2B4*x%2B1%2C+x]
A: The equation has one real solution, but rather ugly I am afraid. 
http://www4f.wolframalpha.com/Calculate/MSP/MSP41323441hc6heb7fh0000004752i2229938d4f0?MSPStoreType=image/gif&s=27&w=286.&h=81.
A: Using maxima (I didn't know maxima could do tex(%)).
$$x=-{{{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}}\over{3\,\left({{
 \sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2}}\right)^{{{1
 }\over{3}}}}}+\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1
 }\over{2}}\right)^{{{1}\over{3}}}\,\left(-{{\sqrt{3}\,i}\over{2}}-{{
 1}\over{2}}\right)-1$$
$$ x=\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}
 }}}+{{1}\over{2}}\right)^{{{1}\over{3}}}\,\left({{\sqrt{3}\,i}\over{
 2}}-{{1}\over{2}}\right)-{{-{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}
 }\over{3\,\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2
 }}\right)^{{{1}\over{3}}}}}-1$$
$$ x=\left({{\sqrt{31}}\over{2\,3^{{{3
 }\over{2}}}}}+{{1}\over{2}}\right)^{{{1}\over{3}}}-{{1}\over{3\,
 \left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2}}\right)
 ^{{{1}\over{3}}}}}-1$$

A: 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 $b$.
