I was looking for the exact solutions of $\cos\frac{2\pi}{257}$, it lead me to the following expressions.
$A=-8\sqrt{514+18\sqrt{257}+12\sqrt{X}+4\sqrt{12593+561\sqrt{257}-98\sqrt{X}-400\sqrt{Y}}}-4\sqrt{514+18\sqrt{257}+12\sqrt{X}-4\sqrt{12593+561\sqrt{257}-98\sqrt{X}-400\sqrt{Y}}}-4\sqrt{514-18\sqrt{257}-12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}-400\sqrt{X}+98\sqrt{Y}}}-4\sqrt{514+18\sqrt{257}-12\sqrt{X}+4\sqrt{12593+561\sqrt{257}+98\sqrt{X}+400\sqrt{Y}}}+\frac{1}{4}*[15+\sqrt{257}+2\sqrt{Y}+2\sqrt{257+15\sqrt{257}+16\sqrt{X}+14\sqrt{Y}}]*\sqrt{514-18\sqrt{257}+12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}$.
with $X=\frac{257+\sqrt{257}}{2}$, $Y=\frac{257-\sqrt{257}}{2}$
Using numerical computation, I noticed that :
$A=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}+4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}$
I did not succeed in doing the demonstration. Do you think it would be possible to demonstrate such a result?
Thank you and good luck.
Close, but not quite. As you defined, let,
$$X =\tfrac{257+\sqrt{257}}{2},\quad\quad Y =\tfrac{257-\sqrt{257}}{2}$$
with your conjugates,
$$A=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}\color{red}{+}4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}=158.3484\dots$$
$$B=6\sqrt{514-18\sqrt{257}+12\sqrt{Y}\color{red}{-}4\sqrt{12593-561\sqrt{257}+400\sqrt{X}-98\sqrt{Y}}}=25.3189 \dots$$
Given the $257$th root of unity $\zeta_{257} = e^{2\pi\,i/257}$. Then,
$$z=\sum_{k=1}^{64} {\zeta_{257}}^{22^k} =-22+\frac{A^2+B^2+12^3\,\sqrt{257}+24^2}{12^3}=9.246073\dots$$
In fact, $z$ is a root of,
$$z^4 + z^3 - 96z^2 - 16z + 256 = 0$$