2014/096) Evaluate $2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$

$2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$
=  $\cos \dfrac{\pi}{7}(2\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}-1)$
= -  $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} -  (2\cos^2 \dfrac{\pi}{7}-1))$
= -  $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} -  \cos \dfrac{2\pi}{7})$
=  -$2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{14}$

=  -$\dfrac{ 2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{14} \cos \dfrac{\pi}{14}}{ \cos \dfrac{\pi}{14}}$
=  -$\dfrac{  \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{7}}{ \cos \dfrac{\pi}{14}}$
=  -$\dfrac{  2\cos \dfrac{\pi}{7}\sin \dfrac{\pi}{7}  \sin\dfrac{3\pi}{14}}{2 \cos \dfrac{\pi}{14}}$
 = -$\dfrac{  \sin  \dfrac{2\pi}{7} \sin\dfrac{3\pi}{14}}{2  \cos \dfrac{\pi}{14}}$
 =- $\dfrac{  \sin  \dfrac{2\pi}{7} \cos (\dfrac{\pi}{2} - \dfrac{3\pi}{14})}{2  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \sin  \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{2  \cos \dfrac{\pi}{14}}$
= - $\dfrac{ 2 \sin  \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{4  \cos \dfrac{\pi}{14}}$
= -$\dfrac{  \sin  \dfrac{4\pi}{7}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \cos(  \dfrac{\pi}{2}- \dfrac{4\pi}{7})}{4  \cos \dfrac{\pi}{14}}$
= -$\dfrac{  \cos  \dfrac{-\pi}{14}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \cos  \dfrac{\pi}{14}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{1}{4}$