prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much. 

By induction?
Well if we assume $3^{3n} + 3^{2n} + 3^n + 1= 4K$ then 
$3^{3(n+1)} +3^{2(n+1)} + 3^{n+1} + 1=$
$3^{3n+3} + 3^{2n + 2} + 3^{n+1} + 1=$
$3^{3n}*27 + 3^{2n}*9 + 3^n*3 + 1 =$
$[3^{3n} + 3^{2n} + 3^n+1] + 26*3^{3n} + 8*3^{2n} + 2*3^n =$
$4K + 4(6*3^{2n} + 2*3^{2n}) + 2(3^{3n} + 3^n) =$
$4[K+(6*3^{2n} + 2*3^{2n})] + 2*3^{n}(3^{2n} + 1)=$
$4[K+(6*3^{2n} + 2*3^{2n})] + 4*3^{n}*\frac {3^{2n} + 1}2$.
The only thing left to prove is that $3^{2n} + 1$ is even and.... c'mon.....!
