$\forall n\in\mathbb N$ prove that at least one of the number $3^{3n}-2^{3n}$,$3^{3n}+2^{3n}$ is divisible by $35$. $3^{3n}-2^{3n}=27^n-8^n=(27-8)(27^{n-1}+27^{n-2}\cdot 8+...+27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even, 
$3^{3n}+2^{3n}=27^n+8^n=(27+8)(27^{n-1}-27^{n-2}\cdot 8+...-27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even and a power of $2$, $3^{3n}+2^{3n}$ can't be factorized.
If $n$ is even, $n=m\cdot 2^k,m>1,k>0$ and $m$ is odd $\Rightarrow$
$$27^n+8^n=(27^{2^k}+8^{2^k})\sum_{i=1}^m 27^{(m-i)2^k}(-b^{2^k})^{i-1}$$
How to check divisibility using these cases?
Reference 

HINT.-Modulo $5$ you have $$3^3\equiv 2\pmod5\\2^3\equiv-2\pmod5$$and módulo $7$ you have $$3^3\equiv -1\pmod7\\2^3\equiv1\pmod7$$
You can deduce easily the result for $3^{3n}\pm2^{3n}$
