Prove that \[ \cos^6x-\sin^6x = \cos2x\left(1-\frac14\sin^22x\right) \]

Proof: Start with the left-hand side: \[ LHS=\cos^6x-\sin^6x \] Using the identity \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] with \[ a=\cos^2x,\quad b=\sin^2x \] we get \[ LHS = (\cos^2x-\sin^2x) (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Using \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[ LHS = \cos2x (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Now, \[ \cos^4x+\sin^4x = (\cos^2x+\sin^2x)^2 – 2\sin^2x\cos^2x \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \cos^4x+\sin^4x = 1-2\sin^2x\cos^2x \] Hence, \[ \cos^4x+\cos^2x\sin^2x+\sin^4x = 1-\sin^2x\cos^2x \] Using \[ \sin2x=2\sin x\cos x \] Squaring both sides: \[ \sin^22x = 4\sin^2x\cos^2x \] Therefore, \[ \sin^2x\cos^2x = \frac14\sin^22x \] Substituting: \[ LHS = \cos2x \left( 1-\frac14\sin^22x \right) \] Hence proved, \[ \boxed{ \cos^6x-\sin^6x = \cos2x\left(1-\frac14\sin^22x\right) } \]