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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.

Rodrigo de Azevedo
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user1094359
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    Is it possible that you came across this problem as an intermediate calculation, when you were trying to solve a different problem? Assuming so, is it possible that you made an arithmetic mistake and that the problem composer of the original problem never intended that you would be confronted with this particular intermediate problem? – user2661923 Oct 22 '22 at 23:12
  • @user2661923 Thanks for comment. No, that is an original problem.Do you know Galois Theory? – user1094359 Oct 22 '22 at 23:33
  • There is certainly exactly one real root, since the polynomial's derivative is everywhere positive. Calculation shows that it's about 0.138. Why do we think it can be expressed via radicals? – Greg Martin Oct 22 '22 at 23:45
  • @GregMartin I think it depends on Galois group. I don't know Galois theory. – user1094359 Oct 22 '22 at 23:54
  • No, I have never studied Galois Theory. – user2661923 Oct 23 '22 at 00:20
  • @GregMartin The Galois group is the Frobenius group of order 42, i.e., the group of affine transformations of $\Bbb F_7$. – Travis Willse Oct 23 '22 at 01:53

2 Answers2

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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.

River Li
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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: $$x_5= 2i\sin\,\left(\frac i7\ln\varphi\right) =2i\cdot\sin\left(i \ln\sqrt[\large7]\varphi\right) =\sqrt[7]\varphi-\frac1{\sqrt[7]\varphi}\approx0.13759740974800.$$

Yuri Negometyanov
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