2024/024) If $\cos\, A +\cos\, B + \cos\, C = 0$, prove that $\cos 3A + \cos 3B +\cos 3C = 12 \cos\, A \cos\, B \cos\, C$.

 We shall use 2 facts

$\cos 3x = 4 \cos^3 x - 3 \cos\, x\cdots(1)$

And 

if $a+b+c=0$ then $a^3+b^3+c^3 = 3abc\cdots(2)$

We are given

 $\cos\, A +\ cos\, B + \cos\, C = 0\cdots(3)$

Hence $\cos^3 A +\ cos^3 B + \cos^3 C = 3 \cos\, A \cos\, B \cos\, C\cdots(4)$

Now $\cos 3A + \cos 3B +\cos 3C$

$= 4 \cos^3 A - 3 \cos\, A$ + $4 \cos^3 B - 3 \cos\, B$ + $4 \cos^3 C - 3 \cos\, C$ using(1)

$=4(\cos^3 A +\ cos^3 B + \cos^3 C - 3 (\cos\, A + \cos\, B + \cos\, C)$

$=4 * 3 \cos\, A \cos\, B \cos\, C- 3 *0$  using (3) and (4)

$=12 \cos\, A \cos\, B \cos\, C$

Proved