Question

if x equals to 2 plus root 3 find x cube plus 1 by x cube

Answer 1

x^3=8+3✓3+12√3+18=26+15√3

(27+15√3)/26+15√3

=[(27+15√3)×(26-15√3)]/1

=702+390√3-405√3-675

=27-15√3

multiply it get it...

brainiest plzzzzzzzzz

Answer 2

Heya there !!!
Here is the answer you were looking for:
$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

On rationalizing the denominator we get,

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

On cubing both the sides we get,

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


Hope this helps!!!

Feel free to ask in comment section if you have any doubt regarding to my answer...

@Mahak24

Thanks...
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