Question
Simplify the following:$\left\{\left( a ^{ m }\right)^{ m -\frac{1}{ m }}\right\}^{\frac{1}{ m +1}}$
✓
Answer
$\left\{\left(a^m\right)^{m-\frac{1}{m}}\right\}^{\frac{1}{m+1}}$
$=(a)^{m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m+1}\right)} \ldots($Using $a^m \div a^n=a^{m-n})$
Consider, $m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=\left(m^2-1\right) \times\left(\frac{1}{m}+1\right)$
$=m^2 \times\left(\frac{1}{m}+1\right)-1 \times\left(\frac{1}{m}+1\right)$
$=\frac{m^2}{m+1}-\frac{1}{m+1}$
$=\frac{m^2-1}{m+1}$
$=\frac{(m-1)(m+1)}{m+1}$
$=m-1(a) m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=a^{m-1} .$
$=(a)^{m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m+1}\right)} \ldots($Using $a^m \div a^n=a^{m-n})$
Consider, $m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=\left(m^2-1\right) \times\left(\frac{1}{m}+1\right)$
$=m^2 \times\left(\frac{1}{m}+1\right)-1 \times\left(\frac{1}{m}+1\right)$
$=\frac{m^2}{m+1}-\frac{1}{m+1}$
$=\frac{m^2-1}{m+1}$
$=\frac{(m-1)(m+1)}{m+1}$
$=m-1(a) m \times\left(m-\frac{1}{m}\right) \times\left(\frac{1}{m}+1\right)$
$=a^{m-1} .$
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