I. Cubic
In 2013, I asked a question regarding an identity by Ramanujan of form,
$$\sqrt[3]{0+2\cos\tfrac{2\pi}{7}}+\sqrt[3]{0+2\cos\tfrac{4\pi}{7}}+\sqrt[3]{0+2\cos\tfrac{6\pi}{7}} = \sqrt[3]{+5-3\,\sqrt[3]{7}}$$
which later turned out to have a hidden partner,
$$\sqrt[3]{1+2\cos\tfrac{2\pi}{7}}+\sqrt[3]{1+2\cos\tfrac{4\pi}{7}}+\sqrt[3]{1+2\cos\tfrac{6\pi}{7}} = \sqrt[3]{-4+3\,\sqrt[3]{7}}$$
There are infinitely many pairs of identities $\sqrt[3]{x_1} + \sqrt[3]{x_2} + \sqrt[3]{x_3}\,$ where $x_i$ are roots of cubics. The natural question is if this has a deg-$5$ version, starting with the minimal polynomial of $2\cos\frac{2\pi}{11}$.
II. Quintic
But the analogy only goes so far: it turns out the cosine expressions under the radical in the LHS now involve products which explains why it was not as easy to find. Let, \begin{align}a &= 2\cos \frac{2\pi}{11},\quad b = 2\cos \frac{4\pi}{11},\quad c = 2\cos \frac{6\pi}{11},\\ d &= 2\cos \frac{8\pi}{11},\quad e= 2\cos \frac{10\pi}{11}\end{align}
We finally have the analogous identity,
$$\sqrt[5]{a\,b^2e^2}+\sqrt[5]{a^2bd^2}+\sqrt[5]{cd^2e^2}+\sqrt[5]{b^2c^2d}+\sqrt[5]{a^2c^2e} = \sqrt[5]{1-5\big(55+21\sqrt[5]{11}-13\sqrt[5]{11^2}-13\sqrt[5]{11^3}\big)}$$
The minpoly of $(a,b,c,d,e)$ is the well-known,
$$u^5 + u^4 - 4u^3 - 3u^2 + 3u + 1 = 0$$
while that of $x_1=ab^2e^2$ (and its conjugates) is the lesser known,
$$x^5 - 6x^4 - x^3 + 32x^2 + 16x + 1 = 0$$
So equivalently,
$$\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}+\sqrt[5]{x_4}+\sqrt[5]{x_5} \\ = \sqrt[5]{1-5\big(55+21\sqrt[5]{11}-13\sqrt[5]{11^2}-13\sqrt[5]{11^3}\big)}$$
The quintic in $x$ is the same given by Noam Elkies (up to sign) in this 2016 MO comment which, after some trial and error, I figured out has factorable roots. And since the quintic in $u$ is a member of the Emma Lehmer family, then by reversing this factorization, this leads to the question.
III. Transformation
The Emma Lehmer quintic is,
$$u^5 + n^2u^4 - (2n^3 + 6n^2 + 10n + 10)u^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)u^2 + (n^3 + 4n^2 + 10n + 10)u +1=0$$
The case $u=2\cos\frac{2\pi}{11}$ is just $n=-1$. Let $x=u_1u_2^2u_5^2$ for the appropriate $u_k$, then we get the new quintic,
$$\frac{(x+1)^5}{(n^4 + 5n^3 + 15n^2 + 25n + 25)} = (n^3 + 3n^2 + 4n + 3) x^4 - (n^4 + 6n^3 + 16n^2 + 23n + 11) x^3 - (n^3 + 5n^2 + 13n + 11) x^2 - (n + 2)x\\$$
The discriminant of both involve the factor $D = (n^4 + 5n^3 + 15n^2 + 25n + 25)^4$. Examples:
Let $n=1, D=71^4$, so $x^5 - 776x^4 + 4057x^3 + 2140x^2 + 218x + 1 = 0$, thus,
$$\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}+\sqrt[5]{x_4}+\sqrt[5]{x_5} = \sqrt[5]{1-5\big(185 - 113 \sqrt[5]{71} - 65 \sqrt[5]{71^2} - \sqrt[5]{71^3} + 12\sqrt[5]{71^4}\big)}$$
Let $n=2, D=191^4$, so $x^5 - 5916x^4 + 35345x^3 + 12425x^2 + 769x + 1 = 0$, thus,
$$\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}+\sqrt[5]{x_4}+\sqrt[5]{x_5} = \sqrt[5]{1-5\big({-299} - 546 \sqrt[5]{191} - 124 \sqrt[5]{191^2} + 20 \sqrt[5]{191^3} + 21 \sqrt[5]{191^4}\big)}$$
and so on.
IV. Questions
Using the roots $x_i$, how do we prove that, $$\sqrt[5]{x_1}+\sqrt[5]{x_2}+\sqrt[5]{x_3}+\sqrt[5]{x_4}+\sqrt[5]{x_5} = \sqrt[5]{\beta}$$ where $\beta$ is just a quintic for general $n$? And how to show $\beta$ has the simple form, $$\beta = c_0+c_1f+c_2f^2+c_3f^3+c_4f^4$$ where $f = (n^4 + 5n^3 + 15n^2 + 25n + 25)^{1/5}$ and $c_i$ are just rationals?
Since $f$ is actually $f = \pm D^{1/20}$ where we used the positive case, does the identity have a partner which uses the negative case (like in the cubic version)?
V. Postscript
With another combination of the Lehmer roots described in this MO post, then,
$$\sqrt[5]{y_1}+\sqrt[5]{y_2}+\sqrt[5]{y_3}+\sqrt[5]{y_4}+\sqrt[5]{y_5} = \color{red}0$$
analogous to orangeskid's cubic version in this MSE post.
