Question

here n= 2^1/3

plzzz solve this question.....

Answer 1

n=2^1/3
=>
$\sqrt[3]{2}$

Answer 2

Use the identity: a^3 -b^3 = (a-b)(a^2 + ab + b^2)

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2}) \\ \\ a - b = \frac{ {a}^{3} - {b}^{3}}{{a}^{2} + ab + {b}^{2}} \\ \\ take \: \: a = n \: \: and \: \: b = 1$

$\frac{n}{n - 1} \\ \\ = \frac{n}{ \frac{ {n}^{3} - {1}^{3} }{ {n}^{2} + n \times 1 + {1}^{2} } } \\ \\ = \frac{n \times ( {n}^{2} + n \times 1 + {1}^{2}) }{{n}^{3} - {1}^{3} } \\ \\ = \frac{n \times ( {n}^{2} + n+1) }{{n}^{3} -1 } \\ \\ =\frac{ {n}^{3} + {n}^{2} +n }{{n}^{3} -1 } \\ \\ = \frac{ {( {2}^{ \frac{1}{3} }) }^{3} + {( {2}^{ \frac{1}{3} }) }^{2} +{2}^{ \frac{1}{3} }}{{( {2}^{ \frac{1}{3} }) }^{3} - 1} \\ \\ = \frac{ 2 + {( {2}^{ \frac{1}{3} }) }^{2} +{2}^{ \frac{1}{3} }}{2 - 1} \\ \\ = 2 + {2}^{ \frac{2}{3} } +{2}^{ \frac{1}{3} }$