Sin^6 альфа+сos^6 альфа+3\4*sin^2 2альфа

Решите задачу:

$\sin^6 \alpha +\cos^6 \alpha +\frac{3}{4}\bullet\sin^22 \alpha =(\sin^2 \alpha)^3 +(\cos^2 \alpha)^3 +\frac{3}{4}\sin^22 \alpha=\\\\=(\frac{1-\cos2 \alpha }{2})^3+(\frac{1+\cos2 \alpha }{2})^3+\frac{3}{4}\sin^22 \alpha =\\\\=\frac{(1-\cos2 \alpha )^3}{8}+\frac{(1+\cos2 \alpha)^3 }{8}+\frac{3}{4}\sin^22 \alpha=\\\\=\frac{1-\cos^32 \alpha-3\cos2 \alpha (1-\cos2 \alpha ) +1+\cos^32 \alpha +3\cos2 \alpha (1+\cos2 \alpha )}{8}+\frac{3}{4}\sin^22 \alpha =$

$=\frac{2-3\cos2 \alpha +3\cos^22 \alpha +3\cos2 \alpha +3\cos^22 \alpha }{8}+\frac{3}{4}\sin^22 \alpha =\\\\=\frac{2+6\cos^22 \alpha }{8}+\frac{3}{4}\sin^22 \alpha=\frac{1+3\cos^22 \alpha }{4}+\frac{3}{4}\sin^22 \alpha =\frac{1+3\cos^22 \alpha +3\sin^22 \alpha }{4}=\\\\=\frac{1+3(\cos^22 \alpha +\sin^22 \alpha )}{4}=\frac{1+3\bullet1}{4}=\frac{4}{4}=1.$