Question

Anyone Please Solve this as soon as possible.​

Answer 1

Question:

If $\sf {a = 2 + \sqrt{3}}$, then find the value of $\sf {a^{3}+ \dfrac {1}{a^{3}}}$.

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Solution:

$\sf {a = 2 + \sqrt{3}}$ [Given]

$\sf {\dfrac {1}{a} = \dfrac{1}{2+ \sqrt{3}}}$

Rationalizing the denominator:

$\sf \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\sf \dfrac{2 - \sqrt{3} }{2 ^{2} - (\sqrt{3})^{2} }$

$\sf \dfrac{2 - \sqrt{3} }{4 - 3 }$

$\sf {2 - \sqrt{3} } = \dfrac{1}{a}$

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Now,

We need to find;

$\sf {a}^{3} + \dfrac{1}{a ^{3} }$

Applying the identity:

(a+b)³ = a³ + b³ + 3ab (a+b)

$\sf ({a + \dfrac{1}{a } }) ^{3} = {a}^{3} + \dfrac{1}{a^{3} } + 3 \times a \times \dfrac{1}{a} (a + \dfrac{1}{a })$

$\sf (2 + \sqrt{3} + 2 - \sqrt{3} ) ^{3} = {a}^{3} + \dfrac{1}{a^{3} } + 3 \times\cancel a \times \cancel \dfrac{1}{{a}} (2 + \sqrt{3} + 2 - \sqrt{3} )$

$\sf (4 ) ^{3} = {a}^{3} + \dfrac{1}{a^{3} } + 3 (4)$

$\sf 64 = {a}^{3} + \dfrac{1}{a^{3} } + 12$

$\sf 64 - 12 = {a}^{3} + \dfrac{1}{a^{3} }$

$\boxed {\red{\bf 52= {a}^{3} + \dfrac{1}{a^{3} }} }$

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Answer:

Value of $\sf {a^{3} + \dfrac {1}{a^{3}}}$ is 52.