Of Cosines and Polynomials

Let $x=2y$ , then $8y^3 - 6y + 1 = 0 \quad\Leftrightarrow\quad 2(4y^3-3y) + 1 = 0$

Recall the triple-angle formula, $\cos(3A) = 4\cos^3(A) - 3\cos(A)$ .

Let $y = \cos(z)$ , then $4y^3-3y = \cos(3z)$ . So, $2\cos(3z) + 1 = 0 \quad\Leftrightarrow\quad \cos(3z) = -\frac12 \quad\Leftrightarrow\quad 3z = \frac23 \pi, \frac43 \pi, \frac83 \pi$

Hence, $x\, = \, 2y \,= \, 2\cos(z) \,= \, 2\cos(\tfrac29 \pi ) , \quad 2\cos(\tfrac49 \pi ) , \quad 2\cos(\tfrac89 \pi )$

The answer is $R+A+B+C+D = 2+2+4+8+9=\boxed{25} .$

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Bonus question: Prove that there exists some cubic polynomial $P$ with integer coefficients whose real root(s) cannot be expressed as $T \cos(U ) + V$ in terms of radicals $T,U$ and $V.$

In equation $x^3-3x+1=0$ we set $x=n \cos α$ to get $n^3\cos^3α-3n\cosα+1=0$

We want this to resemble the trig identity $\cos 3 α = 4 \cos^3 α - 3 \cos α$ . We get the ratio for the coefficients right if we set n=2:

$8 \cos^3 α - 6 \cos α = -1$ while twice the trig identity is $8 \cos^3 α - 6 \cos α = 2 \cos 3 α$

Combine these to get $\cos 3α=-\frac{1}{2}$ $3α = \pm 2π/3 + 2kπ$ $α = \pm 2π/9 + 2kπ/3$ Taking the subset of solutions for which $0<=α<π$ gives $x_1=2\cos \frac {2π}{9},  x_2=2\cos\frac {4π}{9},  x_3=2\cos \frac {8π}{9}$

$R=2, A=2, B=4, C=8, D=9$ Submit the sum $\boxed{25}$

Let $x=R\cos(\theta)$ , then $R^3\cos^3(\theta)-3R\cos(\theta)+1=0 \implies \cos^3(\theta)-\dfrac{3}{R^2}\cos(\theta)+\dfrac{1}{R^3}= 0$ Compare with this Identity $\cos^3(\theta)-\dfrac{3}{4}\cos(\theta)-\cos(3\theta)=0$

We get $R=2$ and $\cos(3\theta)= -\dfrac{1}{2} \implies \theta= \dfrac{2π}{9} ,\dfrac{4π}{9} ,\dfrac{8π}{9}$ So $x= 2\cos(\dfrac{2π}{9}), 2\cos(\dfrac{4π}{9}),2\cos(\dfrac{8π}{9})$

Answer is $\color{#E81990}\boxed{R+A+B+C+D= 25}$