Find the real root of the almost symmetric polynomial $x^7+7x^5+14x^3+7x-1$ 
Find the real root of following almost symmetric polynomial by radicals $$p(x)=x^7+7x^5+14x^3+7x-1$$


Here are my attempts.
The coefficients of $p(x)$ are : $1,7,14,7,-1$.
I wanted to try possible factorizations. But Wolfram Alpha can not factorise this polynomial. This can be a reason of our case, so factorisation over $\Bbb R$ seems impossible.
The Rational root theorem also failed.
Again I tried
$$\begin{align}
x^7+7x^5+14x^3+7x-1
&=x^7+7x^5+7x^3+7x^3+7x-1 \\
&=x^7+7x^3(x^2+1)+7x(x^2+1)-1 \\
&=x^7+7x(x^2+1)^2-1
\end{align}$$
But, this manipulation also didn't work.

Since
$$x^7+7x^5+14x^3+7x=2i\,\text T_7\left(\dfrac{ix}2\right)= 2i \cos\left(7\arccos\,\dfrac{ix}2\right)=1,\tag1$$
where $\;\text T_n(x)\;$ is the Chebyshev polynomial of the first kind (see also WA test),
then
$$x_k=-2i\cos\left(\dfrac17\arccos\,\left(-\dfrac i2\right)+\dfrac{2\pi k}7\right),\qquad(k=0,1,2,3,4,5,6)\tag2$$
(see also WA test),
wherein
$$\arccos\left(-\dfrac i2\right)=-i\ln\left(-\dfrac i2+i\sqrt{1-\left(-\dfrac i2\right)^2}\,\right)=-i\left(\ln e^{^{\Large\frac\pi2i}} + \ln\dfrac1\varphi\right),$$
$$\arccos\left(-\dfrac i2\right)=\dfrac\pi2+i\ln\varphi,\tag3$$
and $\;\varphi=\dfrac{\sqrt5+1}2\;$ is the golden ratio.
From $(2)-(3)$ should
$$x_k=-2i\cos\left(\dfrac{4\pi k+\pi+2i\ln\varphi}{14}\right), \qquad(k=0,1,2,3,4,5,6),\tag4$$
(see also WA test).
If $\,k=5,\,$ then $\dfrac{4k+1}{14}\pi=\dfrac32\pi,$ and we have result in radicals:
$$x_5= 2i\sin\,\left(\frac i7\ln\varphi\right) 
=2i\cdot\sin\left(i \ln\sqrt[\large7]\varphi\right) =\sqrt[7]\varphi-\frac1{\sqrt[7]\varphi}\approx0.13759740974800.$$
