Cosines and polynomials?

This is a way to get the correct answer. However, it does not guarantee that there is a unique triple $(a, b, c)$ . In fact, there is a unique triple, but the proof is too long and complex to be shown here and is left to the reader. Here I present the method to merely get the answer.

Let $y=\frac{720}{7}$ .

Then, note that $\cos (3y)=\cos(0.5y) \implies 4\cos^{3} (y)-3\cos(y)=\pm \sqrt{\frac{\cos(y)+1}{2}}$

(Whether it is plus or minus does not matter as we will square both sides later.)

Now we substitute $x=cos(y)$ to get

$4x^{3}-3x=\pm \sqrt{\frac{x+1}{2}}$

Squaring both sides,

$16x^{6}-24x^{4}+9x^{2}=\frac{x+1}{2}$

Multiplying by 2,

$32x^{6}-48x^{4}+18x^{2}=x+1 \implies 32x^{6}-48x^{4}+18x^{2}-x-1=0$

Hence we get the answer $32+48+18=98.$

$cos7y-1=64(cosy)^7-112cos(y)^5+56cos(y)^3-7cosy-$ as function of cosx.So for angle 720/7 cos(7*720/7)-1=1-1=0 so cos(720/7) is root of 64x^7-112x^5+56x^3-7x-1. Obvious x=1 is another solution,so we can divide by x-1.We now get polynomial 64x^6+64x^5-48x^4-48x^3+8x^2+8x+1.cos(720/7) is also root of this polynomial which is full square of (8x^3+4x^2-4x-1),so cos(720/7) is root of 8x^3+4x^2-4x-1 and that's monic third degree polynomial ith rational coefficiens

(BONUS).Now we just multiply 8x^3+4x^2-4x-1 by (dx^3+ex^2+fx+g) to get polynomial ax^6-bx^4+cx^2-x-1.After calculating we get d=4,e=-2,f=-3,g=1 so (8x^3+4x^2-4x-1)(4x^3-2x^2-3x+1)=32x^6-48x^4+18x^2-x-1 andvalues are a=32,b=48,c=18 ,so a+b+c=32+48+18=98.