$3^{6n}-2^{6n}$

Since $3^{6n} - 2^{6n} \; = \; (3^6)^n - (2^6)^n \; = \; 729^n - 64^n \; = \; (729 - 64)\sum_{j=0}^{n-1}729^{n-j} \times 64^j$ is always, in fact, divisible by $729-64 = 665 = 5\times7\times19$ , it is certainly always divisible by $\boxed{35}$ , but $665$ is the largest possible common divisor (it divides $3^{6n} - 2^{6n}$ for all $n$ , and is equal to $3^{6n} - 2^{6n}$ when $n=1$ ).

3^(6n) - 2^(6n) = (3^6)^n - (2^6)^n = 729^n - 64^n which is divisible by 729 - 64 = 665. The largest factor of 665 given in the options is 35.

Convert in a square minus b square form ,then it will be divisible by 35