Помогите, буду благодарна)(1,2,3)

Решите задачу:

$\displaystyle\tt1) \ \ \frac{8^{11}-8^{10}-8^{9}}{4^{15}-8^{14}-8^{13}}= \frac{8^{9}(8^2-8-1)}{4^{13}(4^2-4-1)}=\frac{{(2^3)}^{9}(64-9)}{{(2^2)}^{13}(16-5)}=\frac{2^{27}\cdot55}{2^{26}\cdot11}=\\\\{} \ \ \ \ =2\cdot5=10 \\\\\\ 2) \ \ \frac{9^{23}-9^{22}-9^{21}}{27^{14}-27^{13}}=\frac{9^{21}(9^{2}-9-1)}{27^{13}(27-1)}=\frac{{(3^2)}^{21}(81-10)}{{(3^3)}^{13}\cdot26}=\frac{3^{42}\cdot71}{3^{39}\cdot26}=\\\\{} \ \ \ \ =\frac{3^3\cdot71}{26}=\frac{27\cdot71}{26}=\frac{1917}{26}=73\frac{19}{26}$

$\displaystyle\tt 3) \ \ \frac{{(8^{n-2}+8^{n-3})}^2}{{(4^{n-1}-4^{n-2})}^3}=\frac{{(8^{n-3}(8+1))}^2}{{(4^{n-2}(4-1))}^3}=\frac{{({(2^3)}^{n-3}\cdot9)}^2}{{({(2^2)}^{n-2}\cdot3)}^3}=\frac{{(2^{3n-9}\cdot3^2)}^2}{{(2^{2n-4}\cdot3)}^3}=\\\\{} \ \ \ \ =\frac{{(2^{3n-9})}^2\cdot{(3^2)}^2}{{(2^{2n-4})}^3\cdot3^3}=\frac{2^{6n-18}\cdot3^4}{2^{6n-12}\cdot3^3}=\frac{3}{2^6}=\frac{3}{64}$