Докажите тождества: 1)(sin³α+cos³α): (sin α+cos α)+sin α·cos α=1 2) sin^4 α+cos^4 α-sin^6 α-cos^6 α=sin²α·cos²α

1) = ( sina + cosa ) ( sin^2a - sina *cosa + cos^2a ) / (sina + cosa) + sina *cosa=
= 1 - sina*cosa + sina* cosa = 1
2)= (sin^2a + cos^2a )^2 - 2sin^2a*cos^2a - ( ( sin^2a)^3 + ( cos^2a)^3 ) =
= 1 - 2 sin^2a*cos^2a - ( sin^2a +cos^2a ) ( (sin^2a)^2 - sin^2a *cos^2a+ (cos^2a)^2)2 ) = 1 - 2 sin^2a*cos^2a -1 + sin^2a*cos^2a= - sin^2a*cos^2a