Prove that \[ \cos4x-\cos4\alpha = 8(\cos x-\cos\alpha) (\cos x+\cos\alpha) (\cos x-\sin\alpha) (\cos x+\sin\alpha) \]

Proof: Start with the left-hand side: \[ LHS=\cos4x-\cos4\alpha \] Using \[ \cos4\theta=8\cos^4\theta-8\cos^2\theta+1 \] we get \[ \cos4x = 8\cos^4x-8\cos^2x+1 \] and \[ \cos4\alpha = 8\cos^4\alpha-8\cos^2\alpha+1 \] Therefore, \[ LHS = (8\cos^4x-8\cos^2x+1) – (8\cos^4\alpha-8\cos^2\alpha+1) \] \[ = 8(\cos^4x-\cos^4\alpha) – 8(\cos^2x-\cos^2\alpha) \] Taking common factor: \[ = 8(\cos^2x-\cos^2\alpha) (\cos^2x+\cos^2\alpha-1) \] Using \[ \sin^2\alpha+\cos^2\alpha=1 \] we get \[ \cos^2x+\cos^2\alpha-1 = \cos^2x-\sin^2\alpha \] Thus, \[ LHS = 8(\cos^2x-\cos^2\alpha) (\cos^2x-\sin^2\alpha) \] Using \[ a^2-b^2=(a-b)(a+b) \] we get \[ \cos^2x-\cos^2\alpha = (\cos x-\cos\alpha) (\cos x+\cos\alpha) \] and \[ \cos^2x-\sin^2\alpha = (\cos x-\sin\alpha) (\cos x+\sin\alpha) \] Therefore, \[ LHS = 8(\cos x-\cos\alpha) (\cos x+\cos\alpha) (\cos x-\sin\alpha) (\cos x+\sin\alpha) \] Hence proved, \[ \boxed{ \cos4x-\cos4\alpha = 8(\cos x-\cos\alpha) (\cos x+\cos\alpha) (\cos x-\sin\alpha) (\cos x+\sin\alpha) } \]