Question

asked Sep 8, 2023 109k views

2 Answers

Mkierc · Answer 1 · 2023-09-10T06:51:49+0000

Given:

$\sf{\dots\implies{\frac{{9}^(n) * {3}^(2) *({{3}^{(- n)/(2) })}^(- 2) -27^2}{ {3}^(3m)*{2}^(3)}}}$

$\rule{80mm}{1pt}$

We need to simply
$\sf{\frac{{9}^(n) * {3}^(2) *({{3}^{(- n)/(2) })}^(- 2) -27^2}{ {3}^(3m)*{2}^(3)}}$ .

$\rule{80mm}{1pt}$

$\sf{\dots\implies{\frac{{9}^(n) * {3}^(2) *({{3}^{(- n)/(2) })}^(- 2) -(27)^2}{ {3}^(3m)*{2}^(3)}}}$

$\sf{\dots\implies{\frac{{3}^(2n) * {3}^(2) *({{3}^{\frac{ \cancel{- }n}{ \cancel2} })}^{ \cancel{- 2}} -(3^(3) )^(2) }{ {3}^(3m)*{2}^(3)}}}$

$\sf{\dots\implies{\frac{{3}^(2n) * {3}^(2) *({{3)}^(n)} -(3 )^(6) }{ {3}^(3m)*{2}^(3)}}}$

Since the base (3) is same so just add the exponents of multiple one.

$\sf{\dots\implies{\frac{{3}^((2n + 2 + n))-(3 )^(6) }{ {3}^(3m)*{2}^(3)}}}$

$\sf{\dots\implies{\frac{{3}^((3n + 2))-(3 )^(6) }{ {3}^(3m)*{2}^(3)}}}$

$\sf{\dots\implies{\frac{{3}^((3n + 2))-(27 )^(2) }{ {3}^(3m)*{2}^(3)}}}$

$\sf{\dots\implies{\frac{{3}^((3n + 2))-( {3}^(2) * {9}^(2) )}{ {3}^(3m)*{2}^(3)}}}$

Take 3² as common.

$\sf{\dots\implies{\frac{ {3}^(2)(({3)}^(3n)-9^(2))}{ {3}^(3m) * 8}}}$

Solve the powers.

$\sf{\dots\implies{\frac{ 9({27}^(n)-81)}{ {27}^(m) * 8}}}$

Again take 27 as common.

$\sf{\dots\implies{\frac{ 9 * 27({1}^(n)-3)}{ {27}^(m) * 8}}}$

Wang Ruiqi · Answer 2 · 2023-09-15T10:16:27+0000

⚘Refer to the attachment for solution and below for steps!!

$\purple{ \rule{300pt}{3pt}}$