$9^7+7^9 is divisible by$

$\begin{array}{rcl}{9}^7+7^9& =& {(8+1)}^7+{(8-1)}^9 \\ & =& {(1+8)}^7-{(1-8)}^9 \\ & =& \left(1+{}^7{\mathrm{C}}_1 \left(8\right)+{}^7{\mathrm{C}}_2 8^2+···\right)-\left(1-{}^9{\mathrm{c}}_1 \left(8\right)+{}^9{\mathrm{c}}_2 8^2-{}^9{\mathrm{C}}_3 8^3+···\right)\\ & =& 8(7+9)+8^2\left({}^7{\mathrm{C}}_2-{}^9{\mathrm{C}}_2\right)+···\\ & =& 8^2 \mathrm{I}\end{array}$

           it is div by 64