The value of $∆ = \left|\begin{array}{ccc}1& \sin 3\theta & {\sin }^3\theta \\ 2\cos \theta & \sin 6\theta & {\sin }^{3}2\theta \\ 4{\cos }^2\theta -1& \sin 9\theta & {\sin }^{3}3\theta \end{array}\right|$ is equal to

Multiplying $C_1$ by $3 \mathrm{s}\mathrm{i}\mathrm{n}$ $\theta$ and dividing $∆$ by $3 \mathrm{s}\mathrm{i}\mathrm{n}$ $\theta$, we get, 

$∆ =\frac{1}{3\sin \theta } \left|\begin{array}{ccc}3\sin \theta & \sin 3\theta & {\sin }^3\theta \\ 3\sin 2\theta & \sin 6\theta & {\sin }^{3}2\theta \\ 3\sin 3\theta & \sin 9\theta & {\sin }^{3}3\theta \end{array}\right|$

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

Now, applying $C_1\leftrightarrow C_1-4C_3$

$C_1$ becomes identical with $C_2$ 

Hence, $∆=0$