$\text{ The value of }\cos {36}^0\cos {42}^0$$\cos {78}^0\text{ is }$$\left[\text{ Given }:\sin 18°=\frac{\sqrt{5}-1}{4}\right.\text{ and }$$\left.\cos 36°=\frac{\sqrt{5}+1}{4}\right]$

$\begin{array}{c}\cos {36}^{°}\cos {42}^{°}\cos {78}^{°}=\cos {36}^{°}\cos \left({60}^{°}-{18}^{°}\right)\cos \left({60}^{°}+{18}^{°}\right)\\ =\frac{\sqrt{5}+1}{4}\left({\cos }^2{60}^{°}-{\sin }^2{18}^{°}\right)\\ ==\left(\frac{\sqrt{5}+1}{4}\right)\left[\frac{1}{4}-{\left(\frac{\sqrt{5}-1}{4}\right)}^2\right]\\ =\left(\frac{\sqrt{5}+1}{4}\right)\frac{1}{4}-\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{5+1-2\sqrt{5}}{16}\right)\\ =\left(\frac{\sqrt{5}+1}{16}\right)-\frac{(\sqrt{5}+1){(\sqrt{5}-1)}^2}{64}\\ =\frac{\sqrt{5}+1}{16}\left[1-\frac{{(\sqrt{5}-1)}^2}{4}\right]\\ =\frac{\sqrt{5}+1}{16}\left[\frac{4-6+2\sqrt{5}}{4}\right]\\ =\frac{1}{8}\end{array}$