Expressing the roots of $y^5 + 11y^4 - 77y^3 + 132y^2 - 77y + 11 = 0$ in terms of of $\zeta_{11}$?

Question

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 $z = \big(\zeta_{11}-\zeta_{11}^{-1}\big)^2$.

Answer 1

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 $u:=t+t^{-1}.$