# 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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