Question

x =
$2 + \sqrt{3}$
Find x⁵ + 1/x⁵​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = 2 + \sqrt{3}$

So, Consider

$\rm :\longmapsto\:\dfrac{1}{x}$

$\rm \:  =  \: \dfrac{1}{2 + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, using this, we get

$\rm \:  =  \: \dfrac{2 - \sqrt{3} }{ {2}^{2} - {( \sqrt{3} )}^{2} }$

$\rm \:  =  \: \dfrac{2 - \sqrt{3} }{4 - 3}$

$\rm \:  =  \: \dfrac{2 - \sqrt{3} }{1}$

$\rm \:  =  \: 2 - \sqrt{3}$

$\rm\implies \:\boxed{\tt{ \frac{1}{x} = 2 - \sqrt{3}}}$

Now, Consider

$\rm :\longmapsto\:x + \dfrac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4$

So, we have

$\rm :\longmapsto\:\boxed{\tt{ x + \dfrac{1}{x} = 4}} - - - (1)$

On squaring both sides, we get

$\rm :\longmapsto\: {\bigg[x + \dfrac{1}{x} \bigg]}^{2} = 16$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 16$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2$

$\rm :\longmapsto\: \boxed{\tt{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 14}} - - - (2)$

Now, Consider

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} }$

$\rm \:  =  \: {\bigg[x + \dfrac{1}{x} \bigg]}^{3} - 3\bigg[x + \dfrac{1}{x} \bigg]$

$\rm \:  =  \: {(4)}^{3} - 3 \times 4$

$\rm \:  =  \: 64 - 12$

$\rm \:  =  \: 52$

So,

$\rm\implies \:\boxed{\tt{ {x}^{3} + \dfrac{1}{ {x}^{3} } = 52}} - - - - (3)$

On multiply equation (2) by (3), we get

$\rm :\longmapsto\:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) \times \bigg( {x}^{3} + \dfrac{1}{ {x}^{3} } \bigg) = 14 \times 52$

$\rm :\longmapsto\: {x}^{5} + \dfrac{1}{x} + x + \dfrac{1}{ {x}^{5} } = 728$

$\rm :\longmapsto\: {x}^{5} + 4 + \dfrac{1}{ {x}^{5} } = 728$

$\rm :\longmapsto\: {x}^{5} + \dfrac{1}{ {x}^{5} } = 728 - 4$

$\rm :\longmapsto\: {x}^{5} + \dfrac{1}{ {x}^{5} } = 724$

Hence,

$\rm\implies \: \underbrace{\boxed{\tt{ \: \: \: \: {x}^{5} + \frac{1}{ {x}^{5} } = 724 \: \: \: \: \: \: }}}$