Question

If
$x - \frac{1}{x} = 1 + 3 \sqrt{2}$,find
$x ^{3} - \frac{1}{{x}^{3} }$
​

Answer 1

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

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

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

${x}^{3} - \frac{1}{ {x}^{3} } = (1 + 3 \sqrt{2} )(1 + 6 \sqrt{2} + 18 +3 )$

${x}^{3} - \frac{1}{ {x}^{3} } = (1 + 3 \sqrt{2} )(22 + 6 \sqrt{2} )$

${x}^{3} - \frac{1}{ {x}^{3} } = 58 + 72 \sqrt{2}$

Answer 2

Answer:

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

Cubing Both The Sides

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

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

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

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

This is the simplified form

On simplifying much Value will come as 159.81