If ${\sin }^{3}x\sin 3x=\sum _{m=0}^n C_m\cos mx\text{ where }{C}_0,C_1,C_2,\dots C_n$ are constants and $C_n\ne 0,\text{ then }n=$

According to the given question we have expressed L.H.S in the form $C_0+C_1\cos x+C_2\cos 2x+\dots +C_n\cos nx$

Now ${\sin }^{3}x\sin 3x=\frac{3\sin x-\sin 3x}{4}\sin 3x$

$\begin{array}{l}=\frac{3\sin x\sin 3x-{\sin }^{2}3x}{4}\\ =\frac{3(\cos 2x-\cos 4x)-(1-\cos 6x)}{8}\end{array}$

Hence $n=6$