Question

plz plz plz plz solve it​

Answer 1

$\large\bf\underline\red{Question \: :- }</h3><h3>$

$\sf{if \: n = 2 + \sqrt{3} \: \: then \:find \: \bigg( n + \frac{1}{n} \bigg) {}^{3} }$

$\large\bf\underline\red{Answer \: :- }$

First we find the value of in bracket then find the cube of the value.

$\sf\longmapsto{ \bigg(2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } \bigg) {}^{3} } \\ \\ \sf\longmapsto{taking \: LCM \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\longmapsto{ \bigg( \frac{(2 + \sqrt{3} )(2 + \sqrt{3} ) + 1}{2 + \sqrt{3} } \bigg) {}^{3} } \\ \\ \sf\longmapsto{ \bigg(\frac{(2 + \sqrt{3} ) {}^{2} + 1}{2 + \sqrt{3} } \bigg) {}^{3}} \\ \\ \sf\longmapsto{ \bigg(\frac{4 + 3+ 1}{2 + \sqrt{3} } \bigg) {}^{3} } \\ \\ \sf\leadsto{ \bigg(\frac{8}{2 + \sqrt{3} } \bigg) {}^{3} }$

Now we find the cube

$\sf\longmapsto{ \frac{512}{8 + \sqrt{27} } } \\ \\ \sf\longmapsto{ \frac{64}{ \sqrt{27} } }$

${\large{\underline{\boxed{\leadsto{\red{ \frac{64}{ \: \sqrt{27} } }}}}}}$

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Answer 2

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

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {\bigg(n + \dfrac{1}{n} \bigg) }^{3}$

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

Given that

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

Consider,

$\bf :\longmapsto\:\dfrac{1}{n}$

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

On rationalizing the denominator, we get

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

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

$\: \: \: \: \: \: \red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

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

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

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

$\bf :\longmapsto\:\dfrac{1}{n} = 2 - \sqrt{3}$

So,

$\rm :\longmapsto\: \bigg(n + \dfrac{1}{n} \bigg)^{3}$

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

$\rm \: = \: \: {4}^{3}$

$\rm \: = \: \:64$

Hence,

$\bf :\longmapsto\: \bigg(n + \dfrac{1}{n} \bigg)^{3} = 64$

Additional Information :-

More Identities to know:

↝(a + b)² = a² + 2ab + b²

↝(a - b)² = a² - 2ab + b²

↝a² - b² = (a + b)(a - b)

↝(a + b)² = (a - b)² + 4ab

↝(a - b)² = (a + b)² - 4ab

↝(a + b)² + (a - b)² = 2(a² + b²)

↝(a + b)³ = a³ + b³ + 3ab(a + b)

↝(a - b)³ = a³ - b³ - 3ab(a - b)