Find the value of $\lambda $, If $\cos A + \cos B + \cos C = 0$ and $\cos 3A + \cos 3B + \cos 3C$$ = $$\lambda \cos A\cos B\cos C$
$
  A.8 \\
  B.10 \\
  C.12 \\
  D.16 \\
$

Hint: In this we use basic identities ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - xz - yz)$ and basic trigonometry expansions ${\cos ^3}(A) + {\cos ^3}(B) + {\cos ^3}(C) = 3\cos A\cos B\cos C$. Use this to find the value of $\lambda $.

Complete step by step answer:
According to the question the equation that is given is $\cos A + \cos B + \cos C = 0........eq1$
We know that ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - xz - yz)$
Let $x = \cos A,y = \cos B$and $z = \cos C$
$\cos A + \cos B + \cos C = x + y + z = 0$,so this identity becomes
$
  {x^3} + {y^3} + {z^3} - 3xyz = 0 \\
  {x^3} + {y^3} + {z^3} = 3xyz \\
$
${\cos ^3}(A) + {\cos ^3}(B) + {\cos ^3}(C) = 3\cos A\cos B\cos C$,
 Now
 $
  \cos (3A) = \cos (2A + A) \\
   = \cos (2A)\cos A - \sin (2A)\sin A \\
   = ({\cos ^2}(A) - {\sin ^2}(A))\cos A - (2\sin A\cos A)\sin A \\
   = {\cos ^3}(A) - 3\cos A{\sin ^2}(A) \\
   = {\cos ^3}(A) - 3\cos A(1 - {\cos ^2}(A)) \\
   = 4{\cos ^3}(A) - 3\cos A \\
$
Similarly $\cos (3B) = 4{\cos ^3}(B) - 3\cos B$and
$\cos (3C) = 4{\cos ^3}(C) - 3\cos (C)$
$
  \therefore \cos (3A) + \cos (3B) + \cos (3C) = 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) \\
   = 12\cos A\cos B\cos C \\
  \therefore \lambda = 12 \\
$
Note: It is advisable to remember such basic identities while involving trigonometric questions as it helps save a lot of time. In the beginning it will be difficult to mug up every identity but with practice things get easier.