Question

please tell the answer​

Answer 1

Given, $\sf {a} = {2 + \sqrt{3} }$

$\sf {\frac{1}{a} = \frac{1}{2 + \sqrt{3} }}$

$\sf {Rationalise\:the\: denominator}$

$\sf{ \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } }$

$\sf{ \frac{1}{a} = \frac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3})}^{2} }}$

$\sf{ \frac{1}{a} = \frac{2 - \sqrt{3} }{ 4 - 3 }}$

$\sf{ \frac{1}{a} = \frac{2 - \sqrt{3} }{1}}$

$\sf{ \frac{1}{a} = {2 - \sqrt{3} }}$

$\sf{ a + \frac{1}{ a} } = 2 + \sqrt{3} + 2 - \sqrt{3}$

$\sf{ a + \frac{1}{ a} } = 4$

$\sf{cubing\:on\:both\:sides,}$

$\sf{{(a + \frac{1}{ a} })}^{3} = {4}^{3}$

$\sf{ {a}^{3} + \frac{1}{ {a}^{3} } + 3(a + \frac{1}{a} ) = 64}$

$\sf{ {a}^{3} + \frac{1}{ {a}^{3} } + 3(4 ) = 64}$

$\sf{ {a}^{3} + \frac{1}{ {a}^{3} } + 12 = 64}$

$\sf{ {a}^{3} + \frac{1}{ {a}^{3} } = 64 - 12}$

$\fbox{\sf{ {a}^{3} + \frac{1}{ {a}^{3} } = 54}}$