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Solution

We need to find: $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$