Prove that:

We know,

$\sin 60^{\circ}=\cos 30^{\circ} \quad\left(=\frac{\sqrt{3}}{2}\right)$

$\Rightarrow \sin 3 \times 20^{\circ}=\cos 3 \times 10^{\circ}$

$\Rightarrow 3 \sin 20^{\circ}-4 \sin ^{3} 20^{\circ}=4 \cos ^{3} 10^{\circ}-3 \cos 10^{\circ}$

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

$\Rightarrow 4\left(\cos ^{3} 10+\sin ^{3} 20^{\circ}\right)=3\left(\cos 10^{\circ}+\sin 20^{\circ}\right)$

Hence proved.