Question

If x=2-root under3, find the value lf (x+1/x)cube+(x+1/x)square+(x+1/x)-100.

Answer 1

Here's your answer !!

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We have to find value of ,

$= > {(x + \frac{1}{x}) }^{3} + {(x + \frac{1}{x}) }^{2} + (x + \frac{1}{x}) - 100 ...(1)$

It's given that,

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

So,

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

By rationalising it ,we get :-

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

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

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

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

By putting value of x and 1/x in equation (1), we get :-

$= > {(2 - \sqrt{3} + 2 + \sqrt{3}) }^{3} + {(2 - \sqrt{3} + 2 + \sqrt{3}) }^{2} + (2 - \sqrt{3} + 2 + \sqrt{3} ) - 100$

$= > ({(4)}^{3} + {(4)}^{2} + 4) - 100 \\ = > (64 + 16 + 4) - 100 \\ = > 84 - 100 \\ = > - 16$

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Hope it helps you !! :)