<s>[INST] Expressing the roots of $y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0$ in terms of of $\zeta_{11}$? Is there a concise way to express the roots of,
$$x^3 + 7x^2 + 7x - 7 = 0$$
using the root of unity $\zeta_7$? 
Similarly, is there an analogous expression for the solvable quintic,
 $$y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0$$
in terms of of $\zeta_{11}$?
P.S. Note the similar-looking,
$$z^5 + 11z^4 + 44z^3 + 77z^2 + 55z + 11 = 0$$
has $z = \big(\zeta_{11}-\zeta_{11}^{-1}\big)^2$.
 [/INST] If $\;t^7=1\;$ and $\;t\ne1,\;$ then $\;x=-1+2/(t+t^{-1})\;$ is a root of $x^3 + 7x^2 + 7x - 7 = 0\;$ but the six primitive $7$th roots come in conjugate pairs giving the three roots of the cubic in $x$.
If $\;t^{11}=1\;$ and $\;t\ne1,\;$ then $\;z=t\!+\!t^{-1}\!-\!2$ is a root of $z^5 + 11z^4 + 44z^3 + 77z^2 + 55z + 11 = 0$ but the ten primitive $11$th roots come in conjugate pairs giving the five roots of the quintic in $z$ and the five roots of $\;y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0\;$ are from $\;y=(2-u)(1-u+u^3)\;$ where $u:=t+t^{-1}.$
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