If ${\sin }^{3}x\sin 3x=\sum _{m=0}^6{c}_m{\cos }^{m}x$ $where$ $c_0,c_1,c_2\mathrm{...}{c}_6$ are constants then                  

$\sum _{m=0}^6{c}_m{\cos }^{m}x={\sin }^{3}x.\sin 3x$

$\Rightarrow c_0+c_1\cos x+c_2{\cos }^{2}x+c_3{\cos }^{3}x+c_4{\cos }^{4}x+c_5{\cos }^{5}x$ $+c_6{\cos }^{6}x={\sin }^{3}x\left(3\sin x-4{\sin }^{3}x\right)$

$=3{\sin }^{4}x-4{\sin }^{6}x$

$=3{\left({\sin }^{2}x\right)}^2-4{\left({\sin }^{2}x\right)}^3$

$=3{\left(1-{\cos }^{2}x\right)}^2-4{\left(1-{\cos }^{2}x\right)}^3$

$=3\left(1+{\cos }^{4}x-2{\cos }^{2}x\right)$ $-4\left(1-{\cos }^{6}x-3{\cos }^{2}x+3{\cos }^{4}x\right)$

$=3+3{\cos }^{4}x-6{\cos }^{2}x-4+4{\cos }^{6}x+12{\cos }^{2}x-12{\cos }^{4}x$

$=-1+6{\cos }^{2}x-9{\cos }^{4}x+4{\cos }^{6}x$

$c_0=-1,c_1=0,c_2=6,c_3=0,c_4=-9,c_5=0,c_6=4$

$c_0+c_2+c_4+c_6=-1+6-9+4=0$