<s>[INST] Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$

The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
 [/INST] $$\frac{a\sqrt{b}+c}{d}~=~2\sqrt{2}~~~~\text{with}~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got 
$$a+d=8~~\text{and}~~\frac{a\sqrt{2}}{d}=2\sqrt{2}$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=\frac{16}3$ and $d=\frac83$. Verifying by setting this values in the first equations:
$$\begin{align}
\frac{\frac{16}{3}\sqrt{2}+0}{\frac83}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16}3+2+0+\frac83~=~10\\
\frac{16\sqrt{2}}{8}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16+8+6}{3}~=~10\\
2\sqrt{2}~=~2\sqrt{2}~~~~&\text{and}~~~~10~=~10
\end{align}$$
Therefore $a=\frac{16}3,b=2,c=0,\frac83$ but this only holds for $a,b,c,d \in \mathbb{Q}$.
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