упростить выражение

$1+\tan^2 \alpha=1/\cos^2 \alpha$

$3-3\cos^2 \alpha=3(1-\cos^2 \alpha)=3\sin^2\alpha$

$\sin^6\alpha+\cos^6\alpha+\dfrac{3-3\cos^2\alpha}{1+\tan^2\alpha}=\sin^6\alpha+(1-\sin^2\alpha)^3+3\sin^2\alpha\cos^2\alpha=$

$=\sin^6\alpha+1-3\sin^2\alpha+3\sin^4\alpha-\sin^6\alpha+3\sin^2\alpha\cos^2\alpha=$

$=1-3\sin^2\alpha(1-\sin^2\alpha)+3\sin^2\alpha\cos^2\alpha=1-3\sin^2\alpha\cos^2\alpha+3\sin^2\alpha\cos^2\alpha=1$