Triggy Radicals

If $z = e^{\frac{2\pi i}{9}}$ then $u = \cos 40^\circ = \tfrac12(z+z^{-1})$ , $v = \cos80^\circ = \tfrac12(z^2 + z^{-2})$ and $w = \cos160^\circ = \tfrac12(z^4 + z^{-4})$ . Note that $z$ is a zero of $0 = X^9-1 = (X^3-1)(X^6 + X^3 + 1)$ , so of $X^6 + X^3 + 1 = 0$ . Thus $\begin{aligned} u+v+w & = \; \tfrac12\big(z + z^{-1} + z^2 + z^{-2} + z^4 + z^{-4}\big) \; = \; -\tfrac12\big(1 +z^3 + z^{-3}\big) \; = \; 0 \\ uv + uw + vw & = \; -\tfrac12(u^2 + v^2 + w^2) \; = \; -\tfrac18(z^2 + 2 + z^{-2} + z^4 +2 + z^{-2} + z + 2 + z^{-1}\big) \; = \; -\tfrac34 \\ uvw & = \; \tfrac18\big(z^3 + z^{-3} + z + z^{-1}\big)\big(z^4 + z^{-4}\big) \; = \; \tfrac18\big(z^2 + z^{-2} + z + z^{-1} + z^4 + z^{-4} + z^3 + z^{-3}\big) \; =\; \tfrac18(z^3 + z^{-3}\big) \; = \; -\tfrac18 \end{aligned}$ so that $u,v,w$ are the roots of the cubic polynomial $8X^3 - 6X + 1 = 0$ .

If we write $\alpha = \sqrt[3]{u}$ , $\beta = \sqrt[3]{v}$ and $\gamma = \sqrt[3]{w}$ then $\begin{aligned} 8X^9 - 6X^3 + 1 & = \; 8(X^3 - u)(X^3 - v)(X^3 - w) \; = \; 8(X - \alpha)(X-\alpha\omega)(X-\alpha\omega^2)(X-\beta)(X-\beta\omega)(X-\beta\omega^2)(X-\gamma)(X-\gamma\omega)(X-\gamma\omega^2) \\ & = \; F(X)F(\omega X)F(\omega^2 X) \end{aligned}$ where $F(X) = 2(X-\alpha)(X-\beta)(X-\gamma) \in \mathbb{R}[X]$ and $\omega=z^3$ is the principal cube root of unity. If we write $F(X) = 2X^3 - aX^2 + bX - c$ for real $a,b,c$ , we deduce that $8X^9 - 6X^3 + 1 \; = \; 8X^9 - \big(a^3 - 6ab + 12c\big)X^6 + \big(b^3 - 3abc + 6c^2\big)X^3 - c^3$ and so that $a^3 - 6ab + 12c \; = \; 0 \hspace{1.5cm} b^3 - 3abc + 6c^2 \; = \; -6 \hspace{1.5cm} c^3 = -1$ so that $c=-1$ and hence $a^3 - 6ab \; = \; 12 \hspace{2cm} b^3 + 3ab \; = \; -12$ Thus $b = \dfrac{a^3 - 12}{6a}$ and hence $\begin{aligned} \left(\dfrac{a^3 - 12}{6a}\right)^3 + 3a\left(\dfrac{a^3 - 12}{6a}\right) + 12 & = \; 0 \\ \big(a^3 - 12\big)^3 + 108a^3\big(a^3 - 12\big) + 2592a^3 & = \; 0 \\ \big(a^3 + 24\big)^3 & = \; 15552 \; = \; 2^6 \times 3^5 \end{aligned}$ Since $a$ is real we deduce that $a^3 = 12\sqrt[3]{9} - 24 = 12\big(\sqrt[3]{9} - 2\big)$ , and hence $\left(\sqrt[3]{\cos40^\circ} + \sqrt[3]{\cos80^\circ} + \sqrt[3]{\cos160^\circ}\right)^3 \; = \; (\alpha + \beta + \gamma)^3 \; = \; \tfrac18a^3 \; = \; \tfrac32\big(\sqrt[3]{9} - 2\big)$ making the answer $9+2=\boxed{11}$ .