Question

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Answer 1

We know ,

(X^n)^m = X^(n.m)

so,

【{(1/x^a^2-b^2)^1/a-b}^a+b】^1/(a+b)^2

= (1/x^a^2-b^2)^ (1/a-b)×(a+b)×1/(a+b)^2

= (1/x^a^2-b^2)^ 1/(a^2-b^2)

= (1/x)^ (a^2-b^2)×1/(a^2-b^2)

= (1/x)^ 1

= 1/x (ans.)