అక్టోబర్
హీట్ అనగానేమి?
Answers
Explanation:
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
Answer:
అక్టోబర్ హిట్ అనగానేమి answer