Question

If $\frac{9^n\times 3^2\times 3^n-(27{)}^n}{3^{3m}\times 2^3}=3^{-3}$, prove that $(m-n)=1$.

Answer 1

Consider given the equation,

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

$\frac{{{9.3}^{2\times }}^n{.3}^n-{\left(3^3\right)}^n}{3^{3m}\times 2^3}=3^{-3}$

$\frac{{9.3}^{3n}-3^{3n}}{3^{3m}\times 2^3}=3^{-3}$

$\frac{{8.3}^{3n}}{3^{3m}\times 2^3}=3^{-3}$

$\frac{3^{3n}}{3^{3m}}=3^{-3}$

$3^{3n-3m}=3^{-3}$

By comparing of power both sides,

$3n-3m=-3$

$n-m=-1$

$m-n=1$

Hence,proved