LHS = sin6x + cos6x
= (sin2x)3 + (cos2x)3
We know that a3 + b3 = (a + b) (a2 + b2 – ab)
= (sin2x + cos2x) [(sin2x)2 + (cos2x)2 – sin2x cos2x]
We know that sin2x + cos2x = 1 and a2 + b2 = (a + b)2 – 2ab
= 1 × [(sin2x + cos2x)2 – 2sin2x cos2x – sin2x cos2x
= 12 - 3sin2x cos2x
= 1 - 3sin2x cos2x = RHS
Hence proved
