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.

For the LHS, use $\cos(120^\circ +A)=\cos 120^\circ \cos A - \sin 120^\circ \sin A$
and $\cos(240^\circ+A)=\cos 240^\circ \cos A - \sin 240^\circ \sin A$
Then $\cos^3(120^\circ+A)=\left(\cos 120^\circ \cos A - \sin 120^\circ \sin A \right)^3$
and $\cos^3(240^\circ+A)=\left( \cos 240^\circ \cos A - \sin 240^\circ \sin A \right)^3$
Pop in values of $\cos 120^\circ$, $\sin 120^\circ$, etc before expanding brackets.
$\cos^3(120^\circ+A)=\left(-1/2 \cos A -\sqrt 3/2 \sin A \right)^3=-\frac 18 \left(\cos A +\sqrt 3 \sin A \right)^3$
and $\cos^3(240^\circ+A)=-\frac 18\left( \cos A - \sqrt3\sin A \right)^3$
