I have tried it with inductive method but can't prove it still. Please help.
Well you should be able to show what you did for the FIRST step of an induction proof. Hard to tell if you are on the right path if you don't show even your first step.
EDIT: Are you required to use induction?
The divisors of 9 are 1, 3, and 9. Saying that "gcd(9, n)= 1" means that n is not a multiple of 3. So n= 3k+ 1 or 3k+ 2. If n= 3k+ 1, you can write this as $\displaystyle (3k+ 1)^6+ 17$ and if n= 3k+ 2 as $\displaystyle (3k+ 2)^6+ 17$ and then do two inductions on k.
Do you really need the inductions?
$\displaystyle \begin{align*} \left( 3\,n + 1 \right) ^6 + 17 &= \left( 3\,n \right) ^6 + 6\,\left( 3\,n \right) ^5 \left( 1 \right) ^1 + 15\,\left( 3\,n \right) ^4 \left( 1 \right) ^2 + 20\,\left( 3\,n \right) ^3\left( 1 \right) ^3 + 15\,\left( 3\,n \right) ^2\left( 1 \right) ^4 + 6\,\left( 3\,n \right) ^1\,\left( 1 \right) ^5 + 1^6 + 17 \end{align*}$
Upon simplification you can clearly see the factor of 9...