Question

if ^ X + 1 is equals to 3 ^ 1 - X then find the value of x​

Answer 1

$\tt GIVEN:-$

$\tt x - \frac{ 1 }{x} = 3$

$\tt FIND:-$

${x}^{3} - \frac{1 }{ {x}^{3} } = ?$

$\tt SOLUTION:-$

$\tt {x}^{3} - \frac{1}{ {x}^{3} }$

$\tt use \: identity \ratio - \\ \tt {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)$

$\tt = > (x - \frac{1}{x } {)}^{3} = {x}^{3} - \frac{1}{ {x}^{3} } - 3 \cancel x \times \frac{1}{ \cancel x} (x - \frac{1}{x} )$

$\tt putting \: x - \frac{1}{x} = 3 \: we \: get$

$\tt = > {3}^{3} = {x}^{3} - \frac{1}{ {x}^{3} } - 3 \times 3$

$\tt = > 27 = {x}^{3} - \frac{1}{ {x}^{3} } - 9$

$\tt = > 27 + 9= {x}^{3} - \frac{1}{ {x}^{3} }$

$\tt = > 36= {x}^{3} - \frac{1}{ {x}^{3} }$

$\tt = > {x}^{3} - \frac{1}{ {x}^{3} } = 36$

$\boxed{ \tt so \: answer \: is \: 36}$