Question

x=2+√3 find the value x³+1/x³​

Answer 1

Given:-

$\\ \sf{ \rightarrow{x = 2 + \sqrt{3} }}$

To Find:-

$\\ \sf{ \rightarrow{ {x}^{3} + \frac{1}{ {x}^{3} } }}$

Formulas Applied:-

$\\ \sf{ \rightarrow{ {a}^{3} + {b}^{3} = (a + b) {}^{3} - 3ab(a + b)}}$

$\sf{ \rightarrow{ {a}^{2} - {b}^{2} = (a + b)(a - b) }}$

Solution:-

First, let's find the value of :-

$\: \sf{ \frac{1}{x} }$

Now, we were given:-

$\\ \sf{x = 2 + \sqrt{3} }$

Therefore,

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

$\\ \sf{ \implies{ \frac{1}{x} } = \frac{(2 - \sqrt{3}) }{(2 + \sqrt{3})(2 - \sqrt{3}) } }$

$\\ \sf{ \implies{ \frac{1}{x} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3} ) {}^{2} } } }\: \: \: \: \: \: \boxed{ \rm{ \because{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}$

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

$\\ \sf{ \therefore{ \bold{ \frac{1}{x} = 2 - \sqrt{3} }}}$

Again,

$\\ \sf{ \bold{x + \frac{1}{x} }}$

$\\ \sf{ \implies{2 + \cancel{\sqrt{3}} + 2 - \cancel{\sqrt{3} }}}$

$\\ \sf{ \implies{ \bold{ 4}}}$

$\\ { \therefore{ \rm{ \pink{x + \frac{1}{x} = 4}}}}$

Now,

$\\ \bold{ {x}^{3} + \frac{1}{ {x}^{3} } }$

$\\ \sf{ \implies{(x) {}^{3} + ( \frac{1}{x} ) {}^{3} }}$

$\\ \sf{ \implies{(x + \frac{1}{x} ) {}^{3} - 3 \times \cancel{ x} \times \frac{1}{ \cancel{x}} (x + \frac{1}{x} })}$

$\\ \sf{ \implies{(4) {}^{3} - 3 \times 4}} \: \: \: \: \: \: \boxed{ \because{ \rm{x + \frac{1}{x} = 4}}}$

$\\ \sf{ \implies{64 - 12}}$

$\\ \sf{ \implies{ \red{52}}}$

$\\ \\ \underline{ \boxed{ \green{ \rm{Hence \: \bold{ {x}^{3} + \frac{1}{ {x}^{3} } = 52}}}}}$

Answer 2

Hope it is helpful for you.