Using binomial theorem, prove that

$3^{2 n+2}-8 n-9=9^{n+1}-8 n-9 \quad \ldots(1)$

Consider

$9^{n+1}=(1+8)^{n+1}$

$\Rightarrow 9^{n+1}={ }^{n+1} C_{0} \times 8^{0}+{ }^{n+1} C_{1} \times 8^{1}+{ }^{n+1} C_{2} \times 8^{2}+{ }^{n+1} C_{3} \times 8^{3}+\ldots+{ }^{n+1} C_{n+1} \times 8^{n+1}$

$\Rightarrow 9^{n+1}=1+8(n+1)+\left[{ }^{n+1} C_{2} \times 8^{2}+{ }^{n+1} C_{3} \times 8^{3}+\ldots+{ }^{n+1} C_{n+1} \times 8^{n+1}\right]$

$\Rightarrow 9^{n+1}-8 n-9=64\left({ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8^{1}+\ldots+{ }^{n+1} C_{n+1} \times 8^{n-1}\right]$

$\Rightarrow 9^{n+1}-8 n-9=64 \times$ An integer

$9^{n+1}-8 n-9$ is divisible by 64

Or,

$3^{2 n+2}-8 n-9$ is divisible by 64 [From (1)]

Hence proved.