Question

Show that :
${\left\{{\left(x^{a-a^1}\right)}^{\frac{1}{a-1}}\right\}}^{\frac{a}{a+1}}=x$

Answer 1

Given,

${\left\{{\left(x^{a-a^{-1}}\right)}^{\frac{1}{a-1}}\right\}}^{\frac{a}{a+1}}$

$={\left\{{\left(x^{\frac{(a+1)(a-1)}{a}}\right)}^{\frac{1}{a-1}}\right\}}^{\frac{a}{a+1}}$

$={\left\{x^{\frac{a+1}{a}}\right\}}^{\frac{a}{a+1}}$

$=x^1=x$