Find the all real roots of the polynomial $x^6+3 x^5+3 x-1=0$ in closed form 
Find the all real roots of the polynomial
$$x^6+3 x^5+3 x-1=0$$
in exact form.

WolframAlpha gives only numerical results. I've asked a few similar questions before. The source of the problem comes from the algebra precalculus workbook (but not homework workbook).
The rational root theorem doesn't work. Because, $x=1$ and $x=-1$ are not roots.
Wolfram says , we have only $2$ real roots. What kind of factorization should I try?Can a factorization of the form $P(x)=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$ work?
I also tried the trick $\frac {P(x)}{x^n}$, but I failed.

This problem is not easy, but you can simplify the work a bit by observing that you can divide both sides by $x^3$. It can be shown that $x=0$ is not a solution to the original polynomial, hence this operation is allowed. We get
$$x^3-\frac 1{x^3}+3\left(x^2+\frac 1{x^2}\right)=0$$
Now, use the two identities below
\begin{align*}
x^3-\frac 1{x^3} & =\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)\\
x^2+\frac 1{x^2} & =\left(x-\frac 1x\right)^2+2
\end{align*}
Our polynomial now becomes
$$\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)+3\left(x-\frac 1x\right)^2+6=0$$
Make the substitution $x-\frac 1x=z$ and the equation is now a cubic in $z$.
$$z^3+3z^2+3z+6=0$$
Use Cardano's method to solve this cubic for $z$ and calculate $x$ by using
$$x-\frac 1x=z$$
The resulting six solutions are the roots of your original polynomial.

EDIT: Here's an actual attempt at solving the problem. Starting with the cubic in $z$, then we get
$$(z+1)^3=-5\qquad\implies\qquad z_0=-1-\sqrt[3]{5}$$
Since the OP is only interested in the real roots, we can neglect the two other complex roots. However, for thoroughness, the two other complex roots would be
\begin{align*}
z_1 & =-1-\omega\sqrt[3]{5}\\
z_2 & =-1-\omega^2\sqrt[3]{5}
\end{align*}
Where $\omega$ is the cube root of unity
$$\omega=\frac {-1+i\sqrt3}2\qquad\qquad\qquad\omega^2=-\frac {1+i\sqrt3}2$$
Returning to the substitution in $x$, we have that
$$x^2-z_0x-1=0$$
A direct application of the quadratic formula gives the two real roots as
\begin{align*}
x_+ & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}+\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)}\\
x_- & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}-\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)}
\end{align*}
