Question

if x=3+√8 find the value of x2+1/x3

Answer 1

Here's your answer!!

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It's given that,

$x = 3 + \sqrt{8}$

We have to find the value of ,

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

Let's find it :-

$x = (3 + \sqrt{8} )$

So,

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} }$

Now,we will rationalise the denominator:-

$= > \frac{1}{x} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} }$

$= > \frac{1}{x} = \frac{3 - \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8} )}^{2} }$

$= > \frac{1}{x} = \frac{3 - \sqrt{8} }{1}$

$= > \frac{1}{x} = 3 - \sqrt{8}$

Now,

We can write,

$(x + \frac{1}{x} ) = ( 3 + \sqrt{8}) - (3 - \sqrt{8} )$

$(x + \frac{1}{x}) = 3 + \sqrt{8} + 3 - \sqrt{8}$

$= > \sqrt{8} - \sqrt{8} \: \: \: get \: cancelled$
$(x + \frac{1}{x}) = 6$

By squaring both side ,we get :-

$= > {( {x} + \frac{1}{x} )}^{2} = {(6)}^{2}$

$= > {(x + \frac{1}{x} )}^{2} = 36$

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We know that,
${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$
We'll applied this identity to solve this question.
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$= > ( {x}^{2} + { \frac{1}{x} }^{2}) + 2 \times x \times \frac{1}{x} = 36$

=>x and x get cancelled.

$= > ( {x}^{2} + { \frac{1}{x} )}^{2} + 2 = 36$

$= > ( {x}^{2} + { \frac{1}{x} )}^{2} = 36 - 2$

$= > ( {x}^{2} + { \frac{1}{x} }^{2} ) = 34$

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Hope it helps you!! :)