$\begin{gathered}
A = 1 - 2\cos x + \cos 2x = \left( {1 + \cos 2x} \right) - 2\cos x = 2{\cos ^2}x - 2\cos x = 2\cos x\left( {\cos x - 1} \right) \hfill \\
B = 1 + \cos x + \cos 2x = 1 + \cos x + 2{\cos ^2}x - 1 = \cos x\left( {2\cos + 1} \right) \hfill \\
C = {\sin ^2}x - {\sin ^2}2x + {\sin ^2}3x = \dfrac{{1 - \cos 2x}}{2} - \dfrac{{1 - \cos 4x}}{2} + \dfrac{{1 - \cos 6x}}{2} = \dfrac{{1 - \cos 2x + \cos 4x - \cos 6x}}{2} = \dfrac{{2{{\sin }^2}x - 2\sin 5x.\sin ( - x)}}{2} \hfill \\
= \dfrac{{2{{\sin }^2}x + 2\sin 5x\sin x}}{2} \hfill \\
= {\sin ^2}x + \sin 5x.\sin x = \sin x\left( {\sin x + \sin 5x} \right) = 2.\sin x\sin 3x.\cos 2x \hfill \\
\end{gathered}$
