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 is such a cursed question.
The only elementary method I can think of is to assume the existence of a factorization of the form $$P(x) = x^6 + 3x^5 + 3x - 1 = (x^2 + kx \pm 1)(x^4 + ax^3 + bx^2 + cx \mp 1) \tag{1}$$ and then after tediously expanding and equating coefficients, we require
$$(a,b,c) = (3-k, 1-3k+k^2, 3-2k+3k^2-k^3) \tag{2}$$ for the case where the quadratic factor is $x^2 + kx - 1$.  For the second case where the quadratic is $x^2 + kx + 1$, the solution for $(a,b,c)$ leads to a contradiction $c = k$ and the linear term becomes $0$; thus is inadmissible.
After substituting $(2)$ into $(1)$ we find
$$P(x) = x^6 + 3x^5 - k g(k) x^2 + (g(k) + 3)x - 1, \tag{3}$$  where $$g(k) = k^3 - 3k^2 + 3k - 6. \tag{4}$$  Then if $g(k) = 0$, we have a valid factorization.  This tells us we are guaranteed a solution in radicals.  We note
$$\begin{align}
g(k) &= k^3 - 3k^2 + 3k - 6 \\
&= (k^3 - 3k^2 + 3k - 1) - 5 \\
&= (k - 1)^3 - 5. \tag{5}
\end{align}$$
Therefore, $g(k) = 0$ has the unique real root $k = 1 + \sqrt[3]{5}$, from which we can refer to $(1)$ to recover the two real roots
$$x = \frac{-k \pm \sqrt{k^2 + 4}}{2} = \frac{-1 - 5^{1/3} \pm \sqrt{5^{2/3} + 2 \cdot 5^{1/3} + 5}}{2}. \tag{6}$$  The quartic does not have any real solutions, but by this point I'm already exhausted by all the tedious math and don't want to prove it by hand.  The problem with this approach is that it is highly speculative--there is no guarantee of it working, and we basically got lucky:  the coefficients in $(3)$ turn out "just right" in which the solution of the cubic $(4)$ makes both the linear and quadratic terms correct.  Moreover, the solution to $(4)$ is "easy" and does not require the cubic formula.
