Question

plz solve it fasttttt
I have exams tomorrow
I need help
if x = 1-√2, then find the value of x³+1/x³

Answer 1

$x + \frac{1}{x} = 1 - \sqrt{2} + \frac{1}{1 - \sqrt{2} } \\ = 1 - \sqrt{2} + \frac{1(1 + \sqrt{2}) }{ {1}^{2} - ({ \sqrt[]{2} })^{2} } \\ = 1 - \sqrt[]{2} - 1 - \sqrt{2} \\ = - 2 \sqrt{2}$
${x}^{3} + ( { \frac{1}{x} })^{3} \\ = ( {x + \frac{1}{x} })^{3} - 3( x + \frac{ 1}{x} ) \\ = ({ - 2 \sqrt{ 2} })^{3} + 6 \sqrt{2} \\ = - 16 \sqrt{2} + 6 \sqrt{2} \\ = - 10 \sqrt{2}$

Answer 2

$Given : x = 1 - \sqrt{2}$

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

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

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

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

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

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Now,

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

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

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On cubing both sides, we get

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

$= > x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = (-16\sqrt{2})$

$= > x^3 + \frac{1}{x^3} + 3(-2\sqrt{2}) = -16\sqrt{2}$

$= > x^3 + \frac{1}{x^3} - 6\sqrt{2} = -16\sqrt{2}$

$= > x^3 + \frac{1}{x^3} = -10\sqrt{2}$

Hope this helps!