Trigonometry+Polynomials=Trigonomials

Let $x = \cos \theta$ , then $a=\cos \theta_1$ , $b=\cos \theta_2$ and $c=\cos \theta_3$ and we have:

$\begin{aligned} 8x^3 - 6x + 1 & = 0 \\ 8\cos^3 \theta - 6 \cos \theta & = -1 \\ 2(4\cos^3 \theta - 3 \cos \theta) & = -1 \\ 2\cos 3\theta & = -1 \\ \cos 3\theta & = - \frac 12 \\ \implies 3\theta & = 120^\circ, \ 240^\circ, \ 480^\circ \\ \theta & = 40^\circ, 80^\circ, 160^\circ \end{aligned}$

Therefore, $\cos^{-1} a + \cos^{-1} b + \cos^{-1} c = 40^\circ + 80^\circ + 160^\circ = \boxed{280}^\circ$

@Chew-Seong Cheong I think we also need to prove $f(x)=8x^3-6x+1$ only have roots lie in $[-1;1]$ so as to set $x=cos(\theta$ )

We have $f'(x)=24x^2-6$ , with $f'(x) =0$ we get $x=\pm\frac{1}{2}$ as roots. We see that the roots of $f(x)=0$ are between $-1$ and $1$ so now we can set $x=cos(\theta)$ and proceed.