Question

(9^n-3^2-3^n-(27)^n) /(3^m*2)^3 =3^-3

Answer 1

$\frac{( {9}^{n} - 3 {}^{2} - {3}^{n} - (27) {}^{2} }{( {3}^{m \times 2) {}^{3} } } = {3}^{ - 3}$

$\frac{( {6}^{n} - {3}^{n} - (729)) }{(6 {m}^{} ) {}^{3} } = {3}^{ - 3}$

$\frac{(3 - 729)}{18m} = - 9$

$\frac{ - 726}{18m} = - 9$

$- 40.33m = - 9$

$m = \frac{ - 40.33}{ - 9}$

$- 49.33$