$If A=6\sin {20}^0–8{\sin }^3{20}^0, B=8 {\cos }^3{20}^0–6\cos {20}^0 and C = \frac{Sin3\theta }{Sin\theta }-\frac{\cos 3\theta }{\cos \theta } then$

$\begin{array}{rcl}\mathrm{A}& =& 6\sin {20}^0-8{\sin }^3{20}^0\\ \mathrm{A}& =& 2(3\sin {20}^0-4{\sin }^3{20}^0)\\ & =& 2\sin {60}^0=\frac{2\sqrt{3}}{2}=\sqrt{3}\\ \mathrm{B}& =& 8{\cos }^3{20}^0-6\cos {20}^0\\ & =& 2(4{\cos }^3{20}^0-3\cos {20}^0)\\ & =& 2\cos {60}^0=2\left(\frac{1}{2}\right)=1\\ \mathrm{C}& =& \frac{\sin 3\mathrm{\theta }}{\mathrm{sin\theta }}-\frac{\cos 3\mathrm{\theta }}{\mathrm{cos\theta }}\\ & =& \frac{\sin 3\mathrm{\theta } \mathrm{cos\theta }-\cos 3\mathrm{\theta } \mathrm{sin\theta }}{\mathrm{sin\theta } \mathrm{cos\theta }}\\ & =& \frac{\sin (3\mathrm{\theta }-\mathrm{\theta })}{\mathrm{sin\theta } \mathrm{cos\theta }}=\frac{\sin 2\mathrm{\theta }}{\mathrm{sin\theta } \mathrm{cos\theta }}\\ & =& \frac{2\mathrm{sin\theta } \mathrm{cos\theta }}{\mathrm{sin\theta } \mathrm{cos\theta }}\\ & =& 2\end{array}$