Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$
There's also another more algebraic way. You can easily show (by expansion) that if $$a+b+c=0$$ then $$a^3+b^3+c^3=3abc$$ Since, in your problem, for every $A$ $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ Then you can use the above identity $$\cos^3{A}+\cos^3{(A+2\pi/3)}+\cos^3{(A-2\pi/3)}=3\cos{A}\cos{(A+2\pi/3)}\cos{(A-2\pi/3)}$$ By using the previously mentioned identity $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos{(x+y)}+\cos{(x-y)}\right)$$ you can simplify the RHS to this $$\frac{3}{2}\cos{A}\left(\cos{(2A)}-\frac{1}{2}\right)$$ and then (using it again) $$\frac{3}{4}\left(\cos{3A}+\cos{(A)}\right)-\frac{3}{4}\cos{A}=\frac{3}{4}\cos{(3A)}$$
You can use this particular formula: $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)$$ twice and simplify $\cos^3{A}$, $\cos^3{(A+2\pi/3)}$, and $\cos^3{(A-2\pi/3)}$ in this way: $$\begin{aligned} \cos^3{A}&=\cos{A}\left(\cos{A}\cos{A}\right) \\ &=\frac{1}{2}\cos{A}\left(1+\cos{2A} \right) \\ &=\frac{1}{2}\cos{A}+\frac{1}{2}\cos{A}\cos{2A}\\ &=\frac{1}{2}\cos{A}+\frac{1}{4}\left(\cos{A}+\cos{3A}\right)\\ &=\frac{1}{2}\cos{A}+\frac{1}{4}\cos{A}+\frac{1}{4}\cos{3A}\\ \end{aligned}$$
And
$$\begin{aligned} \cos^3{(A+2\pi/3)}&=\frac{1}{2}\cos{(A+2\pi/3)}\left(1+\cos{(2A+4\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A+2\pi/3)}\left(1-\cos{(2A+\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{2}\cos{(A+2\pi/3)}\cos{(2A+\pi/3)}\\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\left(\cos{(A-\pi/3)}-\cos{3A}\right)\\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\cos{(A-\pi/3)}+\frac{1}{4}\cos{3A} \end{aligned}$$
And
$$\begin{aligned} \cos^3{(A-2\pi/3)}&=\frac{1}{2}\cos{(A-2\pi/3)}\left(1+\cos{(2A-4\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A-2\pi/3)}\left(1-\cos{(2A-\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{2}\cos{(A-2\pi/3)}\cos{(2A-\pi/3)}\\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\left(\cos{(A+\pi/3)}-\cos{3A}\right)\\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\cos{(A+\pi/3)}+\frac{1}{4}\cos{3A} \end{aligned}$$
Now, by adding the results and using the following identities you can get your answer (the second one can be proved by using the first identity in the reverse order). $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ $$\cos{(A-\pi/3)}+\cos{(A+\pi/3)}=\cos{A}$$
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 $\cos(A+240^\circ) = \cos(360^\circ - (A+240^\circ)) = \cos(120^\circ -A) = \cos(A-120^\circ)$