$a) sin^{6}x + cos^{6}x=(sin^{2}x+cos^{2}x)^{3}-3sin^{2}xcos^{2}x(sin^{2}x+cos^{2}x)=1-3sin^{2}x cos^{2}x=1-\frac{3}{4}sin^{2}2x=\frac{3}{8}(1-2sin^{2}2x)+\frac{5}{8}=\frac{3}{8}\cos 4x+\frac{5}{8}$$b)\sin xcos^{3}x-\cos xsin^{3}x =\sin x\cos x(cos^{2}x-sin^{2}x)=\frac{1}{2}\sin 2x\cos 2x=\frac{1}{4}\sin 4x$