Question

- Simplify :


`\large{ \bf { \frac{( {a}^(2) - \frac{1}{ {b}^(2) } ) ^(a \: ) (a - (1)/(b) ) ^(b - a) }{( {b}^(2) - \frac{1}{ {a}^(2) } ) ^(b) \: (b + (1)/(a) ) \: {}^(a - b) } }}`

Help! Please provide the steps too!
Irrelevant / Random answers will be reported! Ty in advance! :) ​

Answer 1

`\underline{\underline{\red{\textsf {\textbf{ Given :- }}}}}`

• 
  `\sf { \frac{\bigg( {a}^(2) - \frac{1}{ {b}^(2) } \bigg) ^(a \: ) \bigg(a - (1)/(b) \bigg) ^(b - a) }{\bigg( {b}^(2) - \frac{1}{ {a}^(2) } \bigg) ^(b) \: \bigg(b + (1)/(a) \bigg) \: {}^(a - b) } }`

`\underline{\underline{\red{\textsf{\textbf{To \ Find :- }}}}}`

• The simplified form .

`\underline{\underline{\red{\textsf {\textbf{Answer :- }}}}}`

The given expression to us is ,

`\longrightarrow\footnotesize{ \sf \sf { \frac{\bigg( {a}^(2) - \frac{1}{ {b}^(2) } \bigg) ^(a \: ) \bigg(a - (1)/(b) \bigg) ^(b - a) }{\bigg( {b}^(2) - \frac{1}{ {a}^(2) } \bigg) ^(b) \: \bigg(b + (1)/(a) \bigg) \: {}^(a - b) } } }\\\\\\ \longrightarrow\footnotesize{ \sf (\bigg( (a^2b^2-1)/(b^2)\bigg)^a \bigg( (ab-1)/(b)\bigg)^(b-a) )/( \bigg( (b^2a^2-1)/(a^2) \bigg)^b \bigg( ( ba+1)/(a)\bigg)^(a-b) ) } \\\\\\ \longrightarrow\footnotesize{ \sf ( ( ( ab +1)^a(ab-1)^b )/(b^(2a) ) .((ab-1)^(b-a))/(b^(b-a) ))/( ( (ba+1)^b(ba-1)^b )/(a^(2b)). ((ba+1)^(a-b))/(a^(a-b)) ) } \\\\\\ \longrightarrow\footnotesize{ \sf ( ((ab+1)^a ( ab -1)^(a+b-1 ) )/(b^(2a + b - a )))/( ((ab+1)^(b-b + a ) ( ab -1)^b )/( a^(2b + a - b )))} \\\\`

`\\ \longrightarrow\footnotesize{\sf \frac{ \cancel{(ab+1)^a ( ab -1)^b }( a^(a+b)) }{ b^(a + b ) \cancel{( ab +1)^a ( ab -1)^b} } } \\\\`

`\\ \longrightarrow\footnotesize{\sf \boxed{\red{\sf \bigg\lgroup (a)/(b) \bigg\rgroup ^(a+b ) }}}`

Answer 2

• 
  `(ab)^(a+b)`

Explanation:

`(a^2-1/b^2)^a(a - 1/b)^(b-a)/({b^2-1/a^2)^b(b +1/a)^(a-b)) =`

`((a^2b^2-1)/b^2)^a((ab-1)/b)^(b-a)/(((a^2b^2-1)/a^2)^b(ab+1)/a)^(a-b)) =`
`(((ab+1)(ab-1))^a/b^(2a))((ab-1)^(b-a)/b^(b-a))/((((ab+1)(ab-1))^b/a^(2b))(ab+1)^(a-b)/a^(a-b))) =`
`(ab+1)^(a-b-a+b)(ab-1)^(a+b-a-b)b^(2a+b-a)a^(2b+a-b) =`

`(ab+1)^0(ab -1)^0a^(a+b)b^(a+b)=`

`(ab)^(a+b)`