Question

Solve above questions

Answer 1

Here's your answer !!

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1) It is given that,

$\: a \: = 3 - 2 \sqrt{2}$

We have to find value of ,

$= > {a}^{2} - \frac{1}{ {a}^{2} }$

Let's find it ,

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

So,

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

We will now rationalize it :-

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

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

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

So,

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

Now,

$= >( a + \frac{1}{a} ) = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2}$

$- 2 \sqrt{2} \: and \: 2 \sqrt{2} \: \: get \: \: \: cancelled$

$= > (a + \frac{1}{a}) = 6$

And,

$= > (a - \frac{1}{a} ) = 3 - 2 \sqrt{2} - (3 + 2 \sqrt{2} )$

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

$3 \: and \: - 3 \: get \: cancelled$

$= > (a - \frac{1}{a}) = - 4 \sqrt{2}$

We know that,

${a}^{2} + \frac{1}{ {a}^{2} } = (a + \frac{1}{a})(a - \frac{1}{a} )$

$= > 6 \times ( - 4 \sqrt{2} ) \\ = > - 24 \sqrt{2}$

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2) It's given that,

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

We have to find value of ,

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

$if \: x = 2 - \sqrt{3}$

So,

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

We will now rationalize it :-

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

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

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

So,

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

Now,

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

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

$2 \: and \: - 2 \: get \: cancelled$

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

By cubing both side ,we get :-

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

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

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Hope it helps you !! :)