[1-(sinx)^6 -(cosx)^6]/(sin²x-sin⁴x)
=[1-(sin²x+cos²x)(sin⁴x-sin²xcos²x+cos⁴x)]/[sin²x(1-sin²x)]
=(1-sin⁴x+sin²xcos²x-cos⁴x)/(sin²xcos²x)
=[1-(sin²x+cos²x)² +3sin²xcos²x]/(sin²xcos²x)
=(1-1+3sin²xcos²x)/(sin²xcos²x)
=3sin²xcos²x/(sin²xcos²x)
=3