Question

please give me Answer for this question​

Answer 1

Answer:

Option (c) $x^{n-1}$ is the right answer

Step-by-step explanation:

$\huge {\text{ [(x^n)^{n - \frac{1}{n}}]^{\frac{1}{n+1}} }} \\\\\implies \huge {\text{ [(x^n)^{\frac{n^2-1}{n}}]^{\frac{1}{n+1}} }}\\\\\implies \huge {\text{ [(x)^{\frac{n(n^2-1)}{n}}]^{\frac{1}{n+1}} }} \\\\\implies \huge {\text{ [(x)^{{(n^2-1)}}]^{\frac{1}{n+1}} }} \\\\\implies \huge {\text{ [(x)^{{(n+1)(n-1)}}]^{\frac{1}{n+1}} }} \\\\\implies \huge {\text{ (x)^{\frac{(n+1)(n-1)}{n+1}} } \\\\\\\\ \\\implies \huge {(x)^{(n-1)}$

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Answer 2

Answer:

Option (C) is the right answer

Step-by-step explanation:

$[( {x}^{n})^{n - \frac{1}{n} } ]^{ \frac{1}{n + 1} }$

$\scriptsize[( {x}^{n})^{ \frac{n² - 1}{n} } ]^{ \frac{1}{n + 1} }$

$\scriptsize[( {x}^{n})^{ \frac{(n - 1)(n + 1)}{n} } ]^{ \frac{1}{n + 1} } \: [ \because \: {a}^{2} - {b}^{2} = (a - b)(a + b)]$

$( {x}^{n})^{ \frac{(n - 1) \cancel{(n + 1)}}{n} } \times ^{ \frac{1}{ \cancel{(n + 1)}} } \:$

$( {x}^{n} )^{ \frac{(n - 1)}{n} }$

${x}^{ \cancel{n} \times \frac{(n - 1)}{ \cancel{n}} }$

${x}^{n - 1}$