Is there an easy way of showing $8\cos^3(12^o)-6\cos(12^o)=\phi$? Is there an easy way of showing (1)
(1)
$$8\cos^3(12^o)-6\cos(12^o)=\phi$$
with out substituting into the equation?
$8\left(\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\right)^3-6\cdot\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]=\phi$

Trigonometric constant expressed in real radicals-Wikipedia.
$\cos(12^o)=\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]$
$\phi=\frac{1+\sqrt5}{2}$


Recall that $$\cos 3x=4\cos^3 x- 3\cos x$$
$$\implies 2(4\cos^3 12^{\circ}-3\cos 12^{\circ})=2\cos36^\circ$$
Using $$\cos 36^\circ=\frac{1+\sqrt{5}}4$$
We get the that 
$$8\cos^3 12^{\circ}-6\cos 12^{\circ}=\frac{1+\sqrt{5}}2=\phi$$
