For My 25 Followers

Notice that the roots are $x_k=\sqrt[3]{\cos\left(\dfrac{2\pi k}{7}\right)}$ for $k\in\{1,2,3\}$ .

It is a well known fact that $2x_k^3$ are roots of the equation $x^3+x^2-2x-1=0$ , so using Vieta's formula we get $x_1^3+x_2^3+x_3^3=-\frac{1}{2}$ , $(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3=-\frac{1}{2}$ and $(x_1x_2x_3)^3=\frac{1}{8}$ .

Let $S_1=x_1+x_2+x_3$ , $S_2=x_1x_2+x_1x_3+x_2x_3$ and $S_3=x_1x_2x_3=\sqrt[3]{\dfrac{1}{8}}=\dfrac{1}{2}$ .

Cube both sides of the values of $S_1$ and $S_2$ :

$S_1^3=x_1^3+x_2^3+x_3^3+3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-3x_1x_2x_3 \implies S_1^3+2=3S_1S_2 \ldots\ldots (1)$

$S_2^3=(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3+3(x_1x_2+x_1x_3+x_2x_3)(x_1x_2x_3)(x_1+x_2+x_3)-3(x_1x_2x_3)^2 \implies S_2^3+\dfrac{5}{4}=\dfrac{3S_1S_2}{2} \ldots\ldots (2)$

Now substitute $(1)$ in $(2)$ :

$S_2^3+\dfrac{5}{4}=\dfrac{S_1^3+2}{2} \\ S_2=\sqrt[3]{\dfrac{2S_1^3-1}{4}} \ldots\ldots (3)$

Substitute $(3)$ in $(1)$ :

$S_1^3+2=3S_1\sqrt[3]{\dfrac{2S_1^3-1}{4}}$

Cube both sides:

$(S_1^3+2)^3=27S_1^3\left(\dfrac{2S_1^3-1}{4}\right)$

$S_1^9-\dfrac{15}{2}S_1^6+\dfrac{75}{4}S_1^3+8=0$

Note that in this case we can complete the cube to solve for $S_1$ :

$\left(S_1^3-\dfrac{5}{2}\right)^3=-\dfrac{189}{8}$

$S_1=\sqrt[3]{\dfrac{5-3\sqrt[3]{7}}{2}}$

From equation $(1)$ , solve for $S_2$ (we could also solve from equation $(3)$ directly, but it wouldn't be in the required form):

$S_2=\dfrac{S_1^3+2}{3S_1}$

$S_2=\dfrac{3-\sqrt[3]{7}}{2^{2/3}\sqrt[3]{5-3\sqrt[3]{7}}}$

Finally, the equation we want is $x^3-S_1x^2+S_2x-S_3=0$ , so substituting values we get:

$x^3+\left(\sqrt[3]{\dfrac{3\sqrt[3]{7}-5}{2}}\right)x^2+\left(\dfrac{3-\sqrt[3]{7}}{2^{2/3}\sqrt[3]{5-3\sqrt[3]{7}}}\right)x-\dfrac{1}{2}=0$ .

We get $a=2$ , $b=3$ , $c=5$ and $d=7$ . Thus the final answer is $\boxed{17}$ .