Explanation

(b)
We have$,\sin^{3}{x\ \sin\ 3x = \sum_{m = 0}^{n}{c_{m}\cos{mx}}}$
Now$,\sin^{3}{x\sin{3x = \frac{1}{4}(3\sin{x - \sin{3x)\sin{3x}}}}}$
$= \frac{3}{8} \bullet 2\sin{x\sin{3x - \frac{1}{8} \bullet 2\sin^{2}{3x}}}$
$= \frac{3}{8}(\cos{2x - \cos{4x) - \frac{1}{8}(1 - \cos{6x)}}}$
$= - \frac{1}{8} + \frac{3}{8}\cos{2x - \frac{3}{8}\cos{4x + \frac{1}{8}\cos{6x\ \ldots(i)}}}$
$RHS = \sum_{m = 0}^{n}{c_{m}\cos{mx}}$
$= c_{0} + c_{1}\cos{x + c_{2}\cos{2x + c_{3}\cos{3x + ... + c_{n}\cos{nx}}}}$ …(ii)
On comparing Eqs.(i)and (ii), we get$\ n = 6$