1)(1-cos2a)³/8+(1+cos2a)³/8 -1=
(1-3cos2a+3cos²2a-cos³2a+1+3cos2a+3cos²2a+cos³2a-8)/8=
=(6cos²2a-6)/8=6(cos²2a-1)/8=-3sin²2a/4
2)(-3sin²2a/4)³=-27sin^62a/64
3)27(1-cos2a)³(1+cos2a)³/64=27(1-cos²2a³)/64=27sin^62a/64
4)-27sin^62a/64+27sin^62a/64=0
6 - 1 3/5 = 6/1 - 8/5 = 30-8/5 = 22/5 = 4 2/5
Всё подробно написала в решении.