<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] The triangle has area
$$
\frac12 \times \sqrt2 \times (1- \sqrt2/2) = \frac{\sqrt2 - 1}{2},
$$
so the total polygon has area
$$
2\sqrt{2} - \frac{\sqrt2 - 1}{2} = \frac{3}{2}\sqrt{2} + \frac12 = \frac{3\sqrt{2} + 1}{2},
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
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