$\text{ The expression 3 }\left[{\sin }^4\left(\frac{3\pi }{2}-\alpha \right)+{\sin }^4(3\pi +\alpha )\right]-2\left[{\sin }^6\left(\frac{\pi }{2}+\alpha \right)+{\sin }^6(5\pi -\alpha )\right]=$

$\begin{array}{l}{\sin }^4\left(\frac{3\pi }{2}-\alpha \right)={\cos }^4\alpha \\ {\sin }^4(3\pi +\alpha )={\sin }^4\alpha \\ {\sin }^6\left(\frac{\pi }{2}+\alpha \right)={\cos }^6\alpha ,{\sin }^6(5\pi -\alpha )={\sin }^6\alpha \\ \text{Therefore, the given quantity is }=3\left[{\cos }^4\alpha +{\sin }^4\alpha \right]-2\left[{\cos }^6\alpha +{\sin }^6\alpha \right]\\ =3\left[1-2{\sin }^2\alpha {\cos }^2\alpha \right]-2\left[1-3{\sin }^2\alpha {\cos }^2\alpha \right]=1\end{array}$