Trigonometry or elementary algebra?

Let $w=e^{2\pi i/7}$ . Then we know that $w^k+w^{7-k}=2\cos\left(\dfrac{2\pi k}{7}\right)$ . Now let the expression we want be $x$ , and let $a=\cos\left(\dfrac{2\pi}{7}\right)$ , $b=\cos\left(\dfrac{4\pi}{7}\right)$ and $c=\cos\left(\dfrac{6\pi}{7}\right)$ . Hence, $x=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$ .

First we are going to find $a+b+c$ , which is $\dfrac{w+w^6+w^2+w^5+w^3+w^4}{2}=-\dfrac{1}{2}$ . Then $ab+ac+bc$ , which after using the product to sum formulas for cosine we get $-\dfrac{1}{2}$ ; and similarly find $abc$ , which is $\dfrac{1}{8}$ .

Now, cube both sides of the original expression:

$x^3=a+b+c+3(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})(\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc})-3\sqrt[3]{abc}$

Then let $y=\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc}$ :

$x^3=-\dfrac{1}{2}+3xy-\dfrac{3}{2}$

$x^3+2=3xy \ldots\ldots (1)$

Now, cube both sides of the expression with $y$ :

$y^3=ab+ac+bc+3(\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc})(\sqrt[3]{abc})(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})-3\sqrt[3]{(abc)^2}$

$y^3=-\dfrac{1}{2}+\dfrac {3xy}{2}-\dfrac{3}{4}$

$y^3+\dfrac{5}{4}=\dfrac{3xy}{2} \ldots\ldots (2)$

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

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

Finally substitute $(3)$ in $(1)$ :

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

Cube both sides:

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

$x^9+6x^6+12x^3+8=\dfrac{27}{2}x^6-\dfrac{27}{4}x^3$

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

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

$x^9-\dfrac{15}{2}x^6+\dfrac{75}{4}x^3-\dfrac{125}{8}=-\dfrac{125}{8}-8$

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

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

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

On comparing with the given form we get $a=5$ , $b=3$ , $c=7$ and $d=2$ , hence $a+b+c+d=17$ .

let x=cos(2pi/7) , y=cos(4pi/7), z=cos(8pi/7)

Using roots of unity and product to sum formulas you can find that x + y + z = -0.5 xy + yz+ zx = -0.5 xyz = 0.125

Then use Ramanujan's theorem which states if f(x) = x^3 -mx^2 +nx -1 and the roots of f(x) are {a, b, c} then cuberoot(a) + cuberoot(b) + cuberoot(c) = cuberoot(m + 6 +3t) where t^3 - 3(m+n+3)t - (mn+6(m+n)+9)=0

In this case I let the roots of f(x) be 2x, 2y, and 2z because the product of those roots give a constant term of negative one then i divided by the cuberoot of 2 when figuring out cuberoot(x) + cuberoot(y) + cuberoot(z)