$3^{6n}-6^{3n} = \left({3^{2n}}\right)^3 - \left({6^{n}}\right)^3$
$= \left({3^{2n} - 6^n}\right)^3 + 3. 3^{2n}.6^n \left(3^{2n}-6^n\right)$
$= (3^n)^3\left({3^n - 2^n}\right)^3 + 3^{4n+1}. 2^n \left(3^n - 2^n\right)$
$= 3^{3n}\left( {3^n-2^n}\right) \left(\left({3^n-2^n}\right)^2 + 3^{n+1}.2^n\right)$.
So, result is a multiple of$3^{3n}$. Answer is C.
