If $\frac{9^{n} \times 3^{2} \times\left(3^{\frac{-n}{2}}\right)^{-2}-(27)^{n}}{3^{3 m} \times 2^{3}}=\frac{1}{27}$, prove that $m-n=1$
$\frac{9^{n} \times 3^{2} \times\left(3^{\frac{-n}{2}}\right)^{-2}-(27)^{n}}{3^{3 m} \times 2^{3}}=\frac{1}{27}$
$\Rightarrow \frac{\left(3^{2}\right)^{n} \times 3^{2} \times\left(3^{-n}\right)^{-1}-\left(3^{3}\right)^{n}}{3^{3 m} \times 2^{3}}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{2 n} \times 3^{2} \times 3^{n}-3^{3 n}}{3^{3 m} \times 2^{3}}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{2 n+2+n}-3^{3 n}}{3^{3 m} \times 2^{3}}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{3 n+2}-3^{3 n}}{3^{3 m} \times 2^{3}}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{3 n} \times 3^{2}-3^{3 n}}{3^{3 m} \times 2^{3}}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{3 n}(9-1)}{3^{3 m} \times 8}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{3 n}(8)}{3^{3 m} \times 8}=\frac{1}{3^{3}}$
$\Rightarrow \frac{3^{3 n}}{3^{3 m}}=\frac{1}{3^{3}}$
$\Rightarrow 3^{3 n-3 m}=3^{-3}$
$\Rightarrow 3 n-3 m=-3$
$\Rightarrow 3(n-m)=-3$
$\Rightarrow n-m=-1$
$\Rightarrow m-n=1$
Hence, $m-n=1$.
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