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.
 A: 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:
$$x_5= 2i\sin\,\left(\frac i7\ln\varphi\right) 
=2i\cdot\sin\left(i \ln\sqrt[\large7]\varphi\right) =\sqrt[7]\varphi-\frac1{\sqrt[7]\varphi}\approx0.13759740974800.$$
A: Remarks: For a cubic equation $x^3 + px + q = 0$, we use the identity $(u + v)^3 \equiv 3uv(u + v) + u^3 + v^3$ and let $x = u + v$.
Similarly, we have the identities
$$(a+b)^4 - 4ab(a + b)^2 - (a^4 + b^4 - 2a^2b^2) \equiv 0,$$
$$(a+b)^5 - 5ab(a+b)^3 + 5a^2b^2(a+b) - (a^5+b^5) \equiv 0.$$
In general, $a^n + b^n$ ($n\in \mathbb{Z}_{>0}$) can be expressed in terms of $ab$ and $a + b$. See: 1, and 2.

We use the identity
$$(u + v)^7 - 7uv(u+v)^5 + 14u^2v^2(u + v)^3 - 7u^3v^3(u+v) - (u^7+v^7) \equiv 0. \tag{1}$$
Let $x = u + v$. From (1), we have
$$x^7 - 7uvx^5 + 14u^2v^2x^3 - 7u^3v^3x - (u^7+v^7) = 0. \tag{2}$$
If $uv = -1$ and $u^7 + v^7 = 1$, then
(2) gives the equation $x^7+7x^5+14x^3+7x-1 = 0$.
Since $u^7, v^7$ are roots of $y^2 - y - 1 = 0$, we have
$$u = \sqrt[7]{\frac{\sqrt 5  + 1}{2}}, \quad 
v = - \sqrt[7]{\frac{\sqrt 5  - 1}{2}}.$$
Thus, one root of the equation is given by
$$x = u + v = \sqrt[7]{\frac{\sqrt 5  + 1}{2}} - \sqrt[7]{\frac{\sqrt 5  - 1}{2}} \approx 0.1375974100.$$
One can prove that
$$x_k = \mathrm{e}^{\mathrm{i}2\pi k/7} u + \mathrm{e}^{- \mathrm{i}2\pi k/7} v$$
are roots of the equation for $k = 0, 1, \cdots, 6$, which are all roots of the equation.
We are done.
