Question

proove karo point lelo n​

Answer 1

$\huge\green{\mid{\fbox{{คภรฬєг⤵}}\mid}}$

prefer the attachment

$\huge\fbox\red{hσpє}\huge\fbox\pink{ít}\fbox\green{hєlp}\huge\fbox\blue{yóû}$

$\huge\fbox{\red{\underline{mαrk \; mє \; вrαínlíєѕt \; plєαѕє ♥}}}$

$\large\bf{\underline\blue{❥thαnk \; чσu ♥♥}}$

Answer 2

Answer:

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

Given−

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

3

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

ToFind−

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

n

1

)

3

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

Solution−

Given that

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

3

Consider,

\bf :\longmapsto\:\dfrac{1}{n} :⟼

n

1

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

2+

3

1

On rationalizing the denominator, we get

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

2+

3

1

×

2−

3

2−

3

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

2

2

−(

3

)

2

2−

3

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

2

−y

2

}

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

4−3

2−

3

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

1

2−

3

\rm \: = \: \:2 - \sqrt{3} =2−

3

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

n

1

=2−

3

So,

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

n

1

)

3

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

3

+2−

3

)

3

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

3

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

Hence,

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

n

1

)

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)