Доказать 3(sin^4a+cos^4a)-2(sin^6a+cos^a)=1

[latex]3(sin^4 \alpha +cos^4 \alpha )-2(sin^6 \alpha +cos^6 \alpha)=1;[/latex] [latex]3(sin^4\alpha+cos^4\alpha)-2(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)=[/latex] [latex]=3(sin^4\alpha+cos^4\alpha)-2(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)=[/latex] [latex]=3sin^4\alpha+3cos^4\alpha-2sin^4\alpha+2sin^2\alpha cos^2\alpha-2cos^4\alpha=[/latex][latex]=sin^4\alpha+2sin^2\alpha cos^2\alpha+cos^4\alpha=(sin^2\alpha+cos^2\alpha)^2=1;[/latex]

3(sin^4a+cos^4a)-2(sin^6a + cos^6a) =  3(sin^4a+cos^4a)-2(sin^2a+cos^2a)(sin^4a-sin^2a cos^2a+ cos^4a) =  3sin^4a + 3cos^4a- 2sin^4a+2sin^2a cos^2a – 2cos^4a  = sin^4a + 2sin^2a cos^2a+cos^4a = (sin^2a+cos^2a)^2 = 1