If $\displaystyle f\left ( x \right )=\cos ^{2}x+\cos ^{2}\left ( \frac{\pi }{3}+x \right )-\cos x\cos \left ( \frac{\pi }{3} +x\right )$ then $\displaystyle 4f\left ( \dfrac{\pi}8 \right )$ is equal to

$f(x)=\cos ^{ 2 }{ x } +\cos ^{ 2 }{ (\dfrac { \pi }{ 3 } +x) } -\cos { x } \cos { (\dfrac { \pi }{ 3 } +x) } \\ \cos { (a+b) } =\cos { a } \cos { b } -\sin { a } \sin { b } \\ f(x)=\cos ^{ 2 }{ x } +{ \left( \dfrac { 1 }{ 2 } \cos { x } -\dfrac { \sqrt { 3 } }{ 2 } \sin { x } \right) }^{ 2 }-\cos { x } { \left( \dfrac { 1 }{ 2 } \cos { x } -\dfrac { \sqrt { 3 } }{ 2 } \sin { x } \right) }\\ f(x)=\cos ^{ 2 }{ x } +\dfrac { \cos ^{ 2 }{ x } }{ 4 } +\dfrac { 3\sin ^{ 2 }{ x } }{ 4 } -2.\dfrac { 1 }{ 2 } \cos { x } .\dfrac { \sqrt { 3 } }{ 2 } \sin { x } -\dfrac { \cos ^{ 2 }{ x } }{ 2 } +\cos { x } .\dfrac { \sqrt { 3 } }{ 2 } \sin { x } \\ f(x)=\dfrac { 3 }{ 4 } (\cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } )\\ f(x)=\dfrac { 3 }{ 4 } \\ 4f(\dfrac { \pi }{ 8 } )=3$