Answer
The result is $\frac{\sqrt{3}}{2}$.
Work Step by Step
Let us consider the expression:
$\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$
By using the trigonometric identity,
$\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
Now, the above expression can be further simplified as:
$\begin{align}
& \sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)=\sin \left[ \left( \frac{\pi }{3}-\alpha \right)+\left( \frac{\pi }{3}+\alpha \right) \right] \\
& =\sin \left( \frac{\pi }{3}-\alpha +\frac{\pi }{3}+\alpha \right) \\
& =\sin \left( \frac{\pi }{3}+\frac{\pi }{3} \right) \\
& =\sin \left( \frac{2\pi }{3} \right)
\end{align}$
Use, $\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$.
$\sin \left( \frac{2\pi }{3} \right)=\frac{\sqrt{3}}{2}$
Thus, $\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$ can be simplified as $\frac{\sqrt{3}}{2}$.
