Will you find the value of roots?

Consider $z^7=1$ which has roots $z_k=\text{cis}\left(\frac{2 \pi k}{7}\right)$ for $k\in\{0,1,2,3,4,5,6\}$ . Then $z^6+z^5+\cdots+z+1=0$ has roots $z_k=\text{cis}\left(\frac{2 \pi k}{7}\right)$ for $k\in\{1,2,3,4,5,6\}$ . Divide both sides by $z^3$ to get $z^3+\frac{1}{z^3}+z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0$ . With some algebra we get $\left(z+\frac{1}{z}\right)^3+\left(z+\frac{1}{z}\right)^2-2\left(z+\frac{1}{z}\right)-1=0$ . But $y_k=z_k+\frac{1}{z_k}=2\cos\left(\frac{2 \pi k}{7}\right)$ . Note that for the six values of $k$ we only get three different values of $y$ , then the equation in $y$ must be of 3rd degree: $y^3+y^2-2y-1=0$ . Finally make $y=2x$ to obtain $8x^3+4x^2-4x-1=0$ , the answer follows inmediatly and it's $\boxed{9}$ .

As $\color{#20A900}{z^{2n+1}-1\,=\,(z-1) \cdot \large \prod_{k=1}^{n}\left(z^{2}-2 \cdot \cos \frac{2\pi k}{2n+1}z +1 \right) }$ , therefore $z^{7}-1\,=\,z^{2 \cdot 3\,+\,1}-1\,=\,(z-1)\left( z^{2} - 2\cdot \cos\frac{2\pi}{7}\,z + 1\right )\left( z^{2} - 2\cdot \cos\frac{4\pi}{7}\,z + 1\right )\left( z^{2} - 2\cdot \cos\frac{6\pi}{7}\,z + 1\right )$ Also, $\color{#D61F06}{z^{7}-1\,=\,(z-1)(1+z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6})}$ . So, $1+z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}\,=\, \left(z^{2}-2 \cdot \cos\frac{2\pi}{7} z+ 1\right)\left(z^{2}-2 \cdot \cos\frac{4\pi}{7}z + 1\right)\left(z^{2}-2 \cdot \cos\frac{6\pi}{7}z + 1\right)$ On dividing the entire equation by $z^{3}$ , we get :- $\color{magenta}{\left( z^{3} + \frac{1}{z^{3}}\right) \,+\, \left( z^{2} + \frac{1}{z^{2}}\right) \,+\, \left( z + \frac{1}{z}\right) \,+\, 1 \,=\, \left(z-2 \cdot \cos\frac{2\pi}{7} + \frac{1}{z}\right)\left(z-2 \cdot \cos\frac{4\pi}{7} + \frac{1}{z}\right)\left(z-2 \cdot \cos\frac{6\pi}{7} + \frac{1}{z}\right) }$ Say $z+\frac{1}{z}\,=\,t$ , then $z^{2}+\frac{1}{z^{2}}\,=\,t^{2}-2$ and $z^{3}+\frac{1}{z^{3}}\,=\,t^{3}-3t$ . On putting these values in above equation, we get:- $\color{emerald}{t^{3}+t^{2}-2t-1 \,=\, \left(t - 2 \cdot \cos\frac{2\pi}{7}\right)\left(t - 2 \cos\cdot \frac{4\pi}{7}\right)\left(t - 2 \cdot \cos\frac{6\pi}{7}\right)}$ On dividing the entire equation by $8$ , we get :- $\color{#D61F06}{\left(\frac{t}{2}\right)^{3}\,+\, \frac{1}{2} \left(\frac{t}{2}\right)^{2}\,-\,\frac{1}{2}\left(\frac{t}{2}\right)\,-\,\frac{1}{8} \,=\, \left(\frac{t}{2} - \cos\frac{2\pi}{7}\right)\left(\frac{t}{2} - \cos\frac{4\pi}{7}\right)\left(\frac{t}{2} - \cos\frac{6\pi}{7}\right)}$ Say $\frac{t}{2}\,=\,x$ , then we have :- $\color{#20A900}{x^{3}\,+\,\frac{1}{2}x^{2}\,-\,\frac{1}{2}x\,-\,\frac{1}{8} \,=\, \left(x - \cos\frac{2\pi}{7}\right)\left(x -\cos\frac{4\pi}{7}\right)\left(x - \cos \frac{6\pi}{7}\right)}$ . So required equation is :- $\large \color{farkorange}{x^{3}\,+\,\frac{1}{2}x^{2}\,-\,\frac{1}{2}x\,-\,\frac{1}{8} \,=0}$ . $\huge \color{#3D99F6}{\boxed {8x^{3}+4x^{2}-4x-1\,=\,0}}$ . Hence, $\color{#69047E}{a=c=d=2 \,,\, b=3}$ . So, $\color{#69047E}{a+b+c+d\,=\,2+3+2+2\,=\,9}$ .

The Chebyshev Polynomial of the Second Kind $U_6(x)=64x^6-80x^4+24x^2-1$ $=(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)$ has the roots $\cos(\frac{k\pi}{7})$ for $k=1,..,6$ with the first factor giving us the even $k$ . Thus the answer is $2+3+2+2=\boxed{9}$ .

Let

$\kappa = 2\pi/7$ .

Then $3\kappa = 2\pi - 4\kappa$ .

Then $\sin(3\kappa)=-\sin(4\kappa)$ .

Now putting the given values in the above equation at the place of $\kappa$ we find all satisfying.

Hence by using elementary double and triple angle formulas we find

$8x^{3} + 4x^{2} -4x -1$ , where $x=\cos(\kappa)$ .

Hence complaring coefficients we find $a+b+c+d=9$ .

Let $\theta = \frac{2k\pi}{7}$ for $k=1,2,3$ . Then $\cos 4\theta =\cos3\theta$ . This means that $T_4(x)=T_3(x)$ where $x=\cos \theta$ . Now $8x^4-4x^3-8x^2+3x+1=0$ . Factorize it and we have $(x-1)(8x^3+4x^2-4x-1)=0$ .

Now it is clear that $\cos\frac{2k\pi}{7}$ , where $k=1,2,3$ is a root of $8x^3+4x^2-4x-1=0$ . Now $a=2, b=3, c=2, d=2$ and hence $a+b+c+d=\boxed{9}$ .