<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] If $\cos3x=\cos3A$
$3x=n360^\circ\pm3A$ where $n$ is any integer
$x=n120^\circ+ A$ where $n=0,1,2$
As $\cos3x=4\cos^3x-3\cos x,$ the roots of $4c^3-3c-\cos3A=0$ are
$c_{n+1}=\cos(n120^\circ+ A)$ where $n=0,1,2$
Using Vieta's formula, $$c_1+c_2+c_3=0, c_1c_2c_3=\dfrac{\cos3A}4$$  $$\text{ and }c_1c_2+c_2c_3+c_3c_1=-\dfrac34$$
We need $c_1^3+c_2^3+c_3^3$
Now use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
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