Question

[{(1/x^a^2-b^2)^1/a-b}^a+b]^1/(a+b)^2 ​

Answer 1

Answer:

a**2 is the corrrect answer

Answer 2

Answer:

$=\frac{1}{x}$

Step-by-step explanation:

$[{\frac{1}{x^{a^{2} - b^{2} } }^{\frac{1}{a - b} }}^{(a+b)}]^{\frac{1}{(a+b)^{2} } }$

$=[{\frac{1}{x^{a^{2} - b^{2} } }^{\frac{1}{a - b} }}]^{\frac{(a+b)}{(a+b)^{2} } }$

$=[{\frac{1}{x^{a^{2} - b^{2} } }^{\frac{1}{a - b} }}]^{\frac{1}{(a+b) } }$

$=[{\frac{1}{x^{a^{2} - b^{2} } }]^{\frac{1}{a^2 - b^2} }}}$

$=[{\frac{1}{x }]^{\frac{{a^{2} - b^{2} }}{(a^2 - b^2)} }}}$

$=\frac{1}{x}$