Solve it in Friday 13th - part 3

Geometry Level 5

\[ \large \cos\left(\dfrac{2\pi}{13}\right)\cos\left(\dfrac{6\pi}{13}\right)\cos\left(\dfrac{8\pi}{13}\right)\]

If the trigonometric expression above can be expressed as \(\dfrac{a-\sqrt{b}}{c}\) for positive integers \(a,b,c\) with \(b\) is a square-free number and \(\text{gcd}(a,c)=1\), find \(a+b+c\).

See Part 1 and part 2.

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