Math, asked by pranav14161, 1 year ago

if x= 2-root 3 find x -1/x whole cube

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Answered by eshreya396

Here is your answer...
Hope this will help you...

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allysia: go through her steps, they are right

pranav14161: no her answer is not right because 2 and -2 will be cancelled

eshreya396: I have not showed them

eshreya396: but my steps are right

eshreya396: my teacher taught me in this way

pranav14161: OK its right thanks for your help

eshreya396: ok

allysia: no dear, listen.. she has written 2-√2-2-√2 ... Now 2 and -2 are cancelled out and you're left with -√2-√2 and when you add these 2 radicals then they'll be equal to -√2

allysia: hope you aren't confused anymore

allysia: sorry, when yiu add these two radicals then they'll be equal to - 2√2

Answered by allysia

So we have,

$x = 2 - \sqrt{3}$

Using this value of x
you'll get
$\frac{1}{x} = \frac{1}{2 - \sqrt{3} }$

Rationalise the term and you'll get

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

so we got
$\frac{1}{x} = 2 + \sqrt{3}$

Substitute them for the cubic polynomial mentioned above
${((2 - \sqrt{3} ) -( 2 + \sqrt{3})) }^{3} \\ = {(2 - \sqrt{3} - 2 - \sqrt{3} ) }^{3} \\ = { (- 2 \sqrt{3} )}^{3} \\ = - 8 \times 3 \times \sqrt{3} \\ = - 24 \sqrt{3}$

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