Sin^6a+cos^6a+3-3cos^2a/1+tg^2a решите,пожалуйста

Решите задачу:

$sin ^{6} \alpha +cos ^{6} \alpha + \frac{3-3cos ^{2} \alpha}{1+tg ^{2} \alpha }= \\ =(sin ^{2} \alpha)^{3} +(cos ^{2} \alpha)^{3} + \frac{3(1-cos ^{2} \alpha)}{1+ \frac{sin ^{2} \alpha }{cos ^{2} \alpha } }=$
$=(sin^ {2} \alpha +cos ^{2} \alpha)(sin^{4} \alpha -sin^{2} \alpha \cdot cos ^{2} \alpha + cos ^{4} \alpha ) +\frac{3sin ^{2} \alpha\cdot cos ^{2} \alpha }{cos ^{2} \alpha + sin ^{2} \alpha }= \\ = sin ^{4} \alpha +cos ^{4} \alpha -sin^{2} \alpha \cdot cos ^{2} \alpha + 3sin ^{2} \alpha \cdot cos ^{2}\alpha = \\= sin ^{4} \alpha +cos ^{4} \alpha +2sin^{2} \alpha \cdot cos ^{2} \alpha = \\(sin ^{2} \alpha +cos ^{2} \alpha) ^{2} =1$