<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] 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)
$$\frac{3}{4}\left(\cos{3A}+\cos{(A)}\right)-\frac{3}{4}\cos{A}=\frac{3}{4}\cos{(3A)}$$
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