a) 2cosx.cos2x.cos3x - 2sinx.sin2x.sin3x - 1 

= (2cosx.cos3x).cos2x - (2sinx.sin3x).sin2x - 1

=(cos4x + cos2x).cos2x - (cos2x - cos4x).sin2x - 1

$$\begin{gathered} =\cos 4x.\cos 2x +\hspace{0.17em}{\cos }^{2}2x - \cos 2x.\sin 2x +\hspace{0.17em}\cos 4x.\sin 2x - 1 \\ =(\cos 4x.\cos 2x +\hspace{0.17em}\cos 4x.\sin 2x)+({\cos }^{2}2x-1) - \cos 2x.\sin 2x \\ =\cos 4x(\cos 2x + \sin 2x) - {\sin }^{2}2x - \cos 2x.\sin 2x \\ =\cos 4x(\cos 2x +\hspace{0.17em}\sin 2x) - \sin 2x(\sin 2x + \cos 2x) \\ =(\sin 2x +\hspace{0.17em}\cos 2x)(\cos 4x - \sin 2x) \end{gathered}$$

b)

 $$\begin{gathered} 2{\cos }^{3}x+\cos 2x+\sin x \\ =2{\cos }^{3}x+2{\cos }^{2}x - 1 +\hspace{0.17em}\sin x \\ =2{\cos }^{2}x(\cos x +\hspace{0.17em}1) - (1 - \sin x) \\ =2(1 - {\sin }^{2}x)(\cos x + 1) - (1 - \sin x) \\ =(1 - \sin x)\left[2\right(1 +\hspace{0.17em}\sin x\left)\right(\cos x +\hspace{0.17em}1) - 1] \\ =(1 - \sin x)[1 + 2(\sin x + \cos x) + 2\sin x\cos x] \\ =(1 - \sin x)\left[\right(1 + 2\sin x\cos x) + 2(\sin x + \cos x) ] \\ =(1 - \sin x)\left[\right(\sin x +\hspace{0.17em}\cos x{)}^2 + 2(\sin x +\hspace{0.17em}\cos x)] \\ =(1 - \sin x\left)\right(\sin x +\hspace{0.17em}\cos x\left)\right(\sin x + \cos x +\hspace{0.17em}2) \end{gathered}$$