Need Help with Binomial expansion coefficient q

I have to find the coefficient of x^17 for
$\displaystyle \left ( 6x^5-\frac{1}{3x^3} \right )^{21}$

So first I write it out

$\displaystyle \left ( 6x^5-\frac{1}{3x^3} \right )^{21} = \frac{21!}{k!(21-k)!}(6x^5)^{21-k} \left (\frac{-1}{3x^3} \right )^k$

After re arranging I get 105-8k=17 so K isnt a very nice number!

Ive used this method on other expansions and its been fine?

Any help would be appreciated, thank you

your method is correct

you get k=11

*facepalm* I did the difficult bit and messed up with the re arrangement >.<

Thank youu