Let n=3 and θ=4π and you get z2n−2zncosθ+1=z6−z32+1 which you've just shown is expressible as (zn−eiθ)(zn−e−iθ). Putting in values for n and theta gives
z6−z32+1=(z3−ei4π)(z3−e−i4π) so z3=ei4π and z3=e−i4π
Work out the possible values for z and turn them into their cos + i sin forms.