I need the full solution of the questions no. 24 plese help me finding this solution

(sinx+cosx)^2= sin^2x+cos^2x + 2sinxcosx
a^2= 1+2sinx cosx
2sinxcosx= a^2-1............. (1)
now (sin^2x+cos^2x)^3= sin^6x + 3sin^4xcos^2x + 3cos^4xsin^2x + cos^6x
1= sin^6x + cos^6x +3sin^2xcos^2x(sin^2x+cos^2x)
1=sin^6x+cos^6x+3sin^2xcos^2x............(2)
putting the value from (1) in (2), we get
1= sin^6x+cos^6x+ 3/4(a^2-1)^2
therefore sin^6x+cos^6x= 1-3/4(a^2-1)^2............. [hence proved]