We know that
=> sin2A+cos2A=1
=> 1-sin2A=cos2A
=> 1-cos2A=sin2A
Applying these and solving we have:
=2(sin6A+cos6A)-3(sin4A+cos4A)+1
=2sin6A+2cos6A-3sin4A-3cos4A+sin2A+cos2A
=2sin6A+2cos6A-2sin4A-2cos4A-sin4A-cos4A+sin2A+cos2A
=2sin6A-2sin4A+2cos6A-2cos4A+sin2A-sin4A+cos2A-cos4A
=-2sin4A(1-sin2A)-2cos4A(1-cos2A)+sin2A(1-sin2A)+cos2A(1-cos2A)
=-2sin4Acos2A-2cos4Asin2A+sin2Acos2A+cos2Asin2A
=-2sin2Acos2A(sin2A+cos2A)+2sin2Acos2A
=-2sin2Acos2A+2sin2Acos2A
=0
Hence,
2(sin6A+cos6A)-3(sin4A+cos4A)+1=0