MCQ (No Calculators): What is the remainder when dividing $\left \lfloor (6+\sqrt{7})^8 \right \rfloor$ by $9$?

It's not actually an answer, but it's to long for a comment. I am sorry. I tried to use the binomial theorem, but it seems to be very hard. Here is, what i tried.

We have \begin{align*} (6+\sqrt 7)^8&=\sum_{2\mid k}\binom{8}{k}6^{8-k}7^{k/2} +\sqrt 7\cdot \sum_{2\nmid k}\binom{8}{k}6^{8-k}7^{(k-1)/2} \\ &= \sum_{k=0}^4\binom{8}{2k}6^{8-2k}\cdot 7^{k}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\\ &=7^4+\underbrace{6^2\cdot\sum_{k=0}^3\binom{8}{2k}6^{6-2k}\cdot 7^{k}}_{\text{divisible by $9$}}+\sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k \end{align*} One can calulate with much patience, that $$ \lfloor(6+\sqrt 7)^8\rfloor\equiv (-2)^4+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2+\left\lfloor \sqrt 7\cdot \sum_{k=0}^3\binom{8}{2k+1}6^{7-2k}7^k\right\rfloor\equiv -2 + \lfloor 5896848 \sqrt 7\rfloor.$$ Maybe, you can get until here without a calculator. But now I stuck. Sorry sorry