Question

ALERT FOR BRAINLY
$\sqrt{3} + \sqrt{2} find \: {x}^{3} + \frac{1}{ {x}^{3} }$

Answer 1

$Given : x =\sqrt{3} +\sqrt{2}$

$=>\frac{1}{x} =\frac{1}{\sqrt{3} +\sqrt{2}}$

$=>\frac{1}{\sqrt{3} -\sqrt{2}} *\frac{\sqrt{3} -\sqrt{2}} {\sqrt{3} -\sqrt{2}}$

$=>\frac{\sqrt{3} -\sqrt{2}} {1}$

$=>\sqrt{3} -\sqrt{2}$

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Now,

$= > x +\frac{1}{x} =\sqrt{3} +\sqrt{2} +\sqrt{3} -\sqrt{2}$

$=> 2\sqrt{3}$

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On cubing both sides, we get

$= > (x +\frac{1}{x})^3 = (2\sqrt{3})^3$

$=> x^3 +\frac{1}{x^3} + 3(x +\frac{1}{x}) = 24\sqrt{3}$

$=> x^3 +\frac{1}{x^3} + 3(2\sqrt{3}) = 24\sqrt{3}$

$=> x^3 +\frac{1}{x^3} = 24 - 6\sqrt{3}$

$=> x^3 +\frac{1}{x^3} = 18\sqrt{3}$

Therefore,

$=> \boxed{x^3 +\frac{1}{x^3} = 18\sqrt{3}}$

Hope it helps!

Answer 2

Hey mate!!

Given ;

x = √3 + √2

$\frac{1}{x} = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \frac{1}{x} = \frac{ \sqrt{3} - \sqrt{2} }{( \sqrt{3}) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} = \sqrt{3} - \sqrt{2}$

Now,

$x + \frac{1}{x} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} \\ \\ x + \frac{1}{x} = \sqrt{3} + \cancel{ \sqrt[]{2} } + \sqrt{3} - \cancel{ \sqrt[]{2} } \\ \\ x + \frac{1}{x} = 2 \sqrt{3}$

Now,

$= > (x + \frac{1}{x} ) {}^{3} = (2 \sqrt{3} ) {}^{3} \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } + 3(x + \frac{1}{x} ) = 24 \sqrt{3} \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } + 3(2 \sqrt{3} ) = 24 \sqrt{3} \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } = (24 - 6) \sqrt{3} \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } = 18 \sqrt{3}$


Therefore , the Answer is

$\boxed{\bold{\red{\large{x {}^{3} + \frac{1}{x {}^{3} } = 18 \sqrt{3} }}}}$

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THANKS FOR THE QUESTION!

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