Question
Show that:$\Bigg\{\Big(\text{x}^{\text{a}-\text{a}^{-1}}\Big)^{\frac{1}{\text{a}+\text{1}}}\Bigg\}^{\frac{\text{a}}{\text{a}+\text{1}}}=\text{x}$
✓
Answer
$\text{LHS}=\Bigg\{\Big(\text{x}^{\text{a}-\text{a}^{-1}}\Big)^{\frac{1}{\text{a}+\text{1}}}\Bigg\}^{\frac{\text{a}}{\text{a}+\text{1}}}$$=\bigg(\text{x}^{\text{a}-\frac{1}{\text{a}}}\bigg)^{\frac{1}{\text{a}-1}\times\frac{\text{a}}{\text{a}+1}}$
$\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{(\text{a}-1)(\text{a}+1)}}$
$=\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}^{\frac{\text{a}^2-1}{\text{a}}\times\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}$
$=\text{RHS}$
$\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{(\text{a}-1)(\text{a}+1)}}$
$=\bigg(\text{x}^{\frac{\text{a}^2-1}{\text{a}}}\bigg)^{\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}^{\frac{\text{a}^2-1}{\text{a}}\times\frac{\text{a}}{\text{a}^2-1}}$
$=\text{x}$
$=\text{RHS}$
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