Question

.....help me...... ​

Answer 1

We have to prove that :

m - n = 1

$\frac{ {9}^{n} . {3}^{2}. {3}^{n} - {(27)}^{n} }{ {( {3}^{m} .2)}^{3} } = {3}^{ - 3} \\ \\ = > \frac{ {3}^{2n} . {3}^{2} . {3}^{n} - {3}^{3n} }{{3}^{3m}. {2}^{3} } = {3}^{ - 3} \\ \\ = > \frac{1}{8} .( \frac{ {3}^{3n + 2} - {3}^{3n} }{ {3}^{3m} } ) = {3}^{ - 3} \\ \\ = > \frac{1}{8} .( \frac{ {3}^{3n}. {3}^{2} - {3}^{3n} }{ {3}^{3m} } ) = {3}^{ - 3} \\ \\ = > \frac{1}{8} .( \frac{ {3}^{3n} (9 - 1)}{ {3}^{3m} } ) = {3}^{ - 3} \\ \\ = > \frac{ {3}^{3n} }{ {3}^{3m} } = {3}^{ - 3} \\ \\ = > {3}^{3n} . {3}^{ - 3m} = {3}^{ - 3} \\ \\ = > {3}^{3n - 3m} = {3}^{ - 3}$

On comparing :

$3n - 3m = - 3 \\ \\ = > n - m = - 1 \\ \\ = > m - n = 1$

HENCE PROVED .....!!