Question

Show that:

${\displaystyle {[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\}÷\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}]}^{a+b}=1}$

Answer 1

${\displaystyle {[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\}÷\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}]}^{a+b}=1}$

LHS = ${\displaystyle {[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\}÷\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}]}^{a+b}}$

${\displaystyle ={[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\}\times \left\{\frac{x^{b(b+a)}}{x^{b(b-a)}}\right\}]}^{a+b}}$

${\displaystyle ={[\left\{\frac{x^{a^2-ab}}{x^{a^2+ab}}\right\}\times \left\{\frac{x^{b^2+ab}}{x^{b^2-ab}}\right\}]}^{a+b}}$

${\displaystyle ={[\left\{x^{a^2-ab-a^2-ab}\right\}\times \left\{x^{b^2+ab-b^2+ab}\right\}]}^{a+b}}$

${\displaystyle ={[x^{-2ab}\times x^{2ab}]}^{a+b}}$

${\displaystyle ={\left[x^{-2ab+2ab}\right]}^{a+b}}$

${\displaystyle ={\left[x^0\right]}^{a+b}}$

= 1

= RHS