Ta có
$\sin^2a + \sin^2 \left( a - \dfrac{\pi}{3} \right) - \sin a \sin \left( a - \dfrac{\pi}{3} \right)$
$= \dfrac{1}{2} \left[ 1 - \cos(2a) +1 - \cos \left( 2a - \dfrac{2\pi}{3} \right) - 2\sin a \sin \left( a - \dfrac{\pi}{3} \right) \right]$
$= \dfrac{1}{2} \left[2 - \cos \left( 2a - \dfrac{2\pi}{3} \right) - \cos (2a) + \cos \left( 2a - \dfrac{\pi}{3} \right) - \cos \left( \dfrac{\pi}{3} \right) \right]$
$= \dfrac{1}{2} \left[2 +\dfrac{1}{2} . \cos(2a) - sin(2a) . \dfrac{\sqrt{3}}{2} - \cos(2a) + \cos(2a) . \dfrac{1}{2} + \sin(2a) . \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} \right]$
$= \dfrac{1}{2} \left( 2 - \dfrac{1}{2}\right)$
$= \dfrac{1}{2} . \dfrac{3}{2} = \dfrac{3}{4}$
Vậy ko phụ thuộc vào $a$.