<s>[INST] Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:

$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$

My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
 [/INST] Consider the following picture. We see that the triangle with vertices $(\cos A, \sin A), (\cos(A+120^\circ), \sin(A+120^\circ)), (\cos(A+240^\circ), \sin(A+240^\circ))$ are the vertices of an equilateral triangle with centriod at the origin.

Thus if we set $a = \cos A, b = \cos(A+120^\circ), c = \cos(A+240^\circ)$, then
\begin{align*}
a+b+c = 0
\end{align*}
and hence 
\begin{align*}
a^3+b^3+c^3 = 3abc
\end{align*}
and 
\begin{align*}
\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)} &= 3\cos A \cos(A+120^\circ)\cos(A+240^\circ) \\
&= \frac{3}{2}(2\cos A \cos(A+120^\circ))\cos(A+240^\circ) \\
&= \frac{3}{2}[\cos(2A+120^\circ)+\cos(120^\circ)]\cos(A+240^\circ)\\
&=\frac{3}{2}\cos(2A+120^\circ)\cos(A+240^\circ)-\frac{3}{4}\cos(A+240^\circ)\\
&=\frac{3}{4}\cos(3A+360^\circ)+\frac{3}{4}\cos(A-120^\circ) -\frac{3}{4}\cos(A+240^\circ)\\
&=\frac{3}{4}\cos(3A)
\end{align*}
since $\cos(A+240^\circ) = \cos(360^\circ - (A+240^\circ)) = \cos(120^\circ -A) = \cos(A-120^\circ)$
</s>