Question

$if \: x - \frac{1}{x } = 3 + 2 \sqrt{2} \: \: find \: the \: value \: of \: {x}^{3} - \frac{1}{{x}^{3} }$
​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf\red{ {x}^{3} - \frac{1}{ {x}^{3} } = 108 + 76 \sqrt{2}}}$

$\mathfrak{\large{\underline{\underline{Given:-}}}}$

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

$\mathfrak{\large{\underline{\underline{To find:-}}}}$

the value of ${x}^{3} - \frac{1}{{x}^{3} }$

$\mathfrak{\large{\underline{\underline{Solution:-}}}}$

Cubing on both sides we get,

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

$\implies$ $\bold{ {x}^{3} - \frac{1}{ {x}^{3} } -3 x \times \frac{1}{x} (x - \frac{1}{x}) = {3}^{3} + (2 \sqrt{2} )^{3} + 3. {3}^{2}.2 \sqrt{2} + 3.3 {(2 \sqrt{2} )}^{2} .}$

put the value of x - 1/x

$\implies$ $\bold{ {x}^{3} - \frac{1}{ {x}^{3} } - 3.(3 + 2 \sqrt{2} ) = 27 + 16 \sqrt{2} + 54 \sqrt{2} + 72}$

$\implies$ $\bold{ {x}^{3} - \frac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} = 99 + 70 \sqrt{2} }$

$\implies$ $\bold{ {x}^{3} - \frac{1}{ {x}^{3} } = 99 + 9 + 70 \sqrt{2} + 6 \sqrt{2} }$

$\implies$ $\bold{{x}^{3} - \frac{1}{ {x}^{3} } = 108 + 76 \sqrt{2} }$

Identity used :-

$\boxed</strong><strong>{\sf</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>(x - y) ^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)</strong><strong> }}$

$\boxed{\sf</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{ </strong><strong>(x - y) ^{3} = {x}^{3} - {y}^{3} - 3 {x}^{2} y - 3x {y}^{2} </strong><strong>}}$