Question

if x=2-√3 find the value of x3+1/x3

Answer 1

Hey mate !

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Given :

$x = 2 - \sqrt{3}$

To find :

$x {}^{3} + \frac{1}{x{}^{3}}$

Solution :

$x = 2 - \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 + \sqrt{3}$

Now,

$x + \frac{1}{x} = 2 - \cancel {\sqrt{3}} + 2 + \cancel {\sqrt{3}} \\ \\ x + \frac{1}{x} = 2 + 2 \\ \\ x + \frac{1}{x} = 4$

And,

On cubing both sides.

$(x + \frac{1}{x} ){}^{3} = (4) {}^{3} \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 3 \times 4 = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 12 = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } = 64 - 12 \\ \\ \boxed{ \bold{ x {}^{3} + \frac{1}{x{}^{3} } = 52}}$

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Thanks for the question!

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Answer 2

$\underline{\bold{Given:-}}$

$x = 2 - \sqrt{3}$

$\underline{\bold{To\:find:-}}$

${x}^{3} + \frac{1}{ {x}^{3} }$

$\underline{\bold{Solution:-}}$

$x = 2 - \sqrt{3}$

$\frac{1}{x} = \frac{1}{2 - \sqrt{3} } \\ \\ \bold{On \: rationalising \: them }\\ \\ \frac{1}{x} = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{ {2}^{2} - {( \sqrt{3}) }^{2} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 + \sqrt{3} \: \: \: \: \: .......(1)$
$x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ \\ x + \frac{1}{x} = 4 \: \: \: \: \: .......(2) \\ \\ \bold{On \: cubing \: both \: sides }\\ \\ \bold{{(x + \frac{1}{x})}^{3} = {4}^{3} }$

${x}^{3} + \frac{1}{ {x}^{3} } + 3x \times \frac{1}{x} (x + \frac{1}{x} ) = 64 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ \\ from \: eq \: (2) \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(4) = 64 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \\\boxed{\bold{ {x}^{3} + \frac{1}{ {x}^{3} } = 52}}$

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