Prove that \[ \cos^32x+3\cos2x=4(\cos^6x-\sin^6x) \]

Proof: Start with the right-hand side: \[ RHS=4(\cos^6x-\sin^6x) \] Using the identity \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] we get \[ \cos^6x-\sin^6x = (\cos^2x-\sin^2x) (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Using \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[ RHS = 4\cos2x (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] 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, \[ \sin^2x\cos^2x=\frac{\sin^22x}{4} \] Substituting: \[ RHS = 4\cos2x \left( 1-\frac{\sin^22x}{4} \right) \] \[ = 4\cos2x-\cos2x\sin^22x \] Using \[ \sin^2\theta=1-\cos^2\theta \] we get \[ RHS = 4\cos2x-\cos2x(1-\cos^22x) \] \[ = 4\cos2x-\cos2x+\cos^32x \] \[ = 3\cos2x+\cos^32x \] \[ = \cos^32x+3\cos2x \] \[ =LHS \] Hence proved, \[ \boxed{ \cos^32x+3\cos2x = 4(\cos^6x-\sin^6x) } \]