Binomial theorem for positive integral Index

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The remainder when $8^{2 n} - 62^{2 n + 1}$ is divided by $9$ is

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$8^{2 n} - 62^{2 n + 1}$
$\begin{array}{l} = (9 - 1)^{2 n} - (63 - 1)^{2 n + 1} \\ = (^{2 n} C_{0} 9^{2 n_{-} 2 n} C_{1} 9^{2 n - 1} +^{2 n} C_{2} 9^{2 n - 2} - \dots - \\ ^{2 n} C_{2 n - 1} (9) +^{2 n} C_{2 n}) - (^{2 n + 1} C_{0} (63^{2 n + 1}) -^{2 n + 1} C_{1} (63^{2 n}) \\ +^{2 n + 1} C_{2} (63^{2 n - 1}) - \dots - +^{2 n + 1} C_{1} (63) -^{2 n + 1} C_{0}) \end{array}$
$= 9 m + 1 + 1 = 9 m + 2$ for some integer $m$
Thus, when $3^{2 n} - 62^{2 n + 1}$
is divided by, $9$ the remainder is $2$ .

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Binomial theorem for positive integral Index

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The remainder when 82n−622n+1is divided by 9 is

82n−622n+1=(9−1)2n−(63−1)2n+1= 2nC092n−2nC192n−1+2nC292n−2−⋯− 2nC2n−1(9)+2nC2n− 2n+1C0632n+1−2n+1C1632n+2n+1C2632n−1−⋯−+2n+1C1(63)−2n+1C0=9m+1+1=9m+2 for some integer m Thus, when 32n−622n+1is divided by, 9 the remainder is 2.

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The remainder when $8^{2 n} - 62^{2 n + 1}$ is divided by $9$ is