Question

If x-1 / x cube = 3 + 2 root2, find the
value of x cube-1/x cube​

Answer 1

$\huge\sf\underline { Question } :$

• $\sf \: If \: \: {x}^{3} + \dfrac{ 1 }{ {x}^{3} } = 3 + 2 \sqrt{2} \: \: then \: find \: the \\ \: \sf value \: of \: \bigg( x - \dfrac{ 1 }{ x } \bigg ) .$

$\huge\sf\underline { To Find } :$

• Find the value.

$\huge\sf\underline { Solution } :$

$\longrightarrow\sf \: { \bigg( x - \dfrac{ 1 }{ x } \bigg ) }^{ 3 } = {x}^{3} + \dfrac{ 1 }{ {x}^{3} } - 2 \times x \times \dfrac{1}{x}$

$\longrightarrow\sf \: { \bigg( x - \dfrac{ 1 }{ x } \bigg ) }^{ 3 } = 3 + 2 \sqrt{2} - 2$

$\longrightarrow\sf { \bigg( x - \dfrac{ 1 }{ x } \bigg ) }^{ 3 } = 1 + 2 \sqrt{2}$

$\longrightarrow\sf \bigg( x - \dfrac{ 1 }{ x } \bigg ) = \sqrt{1 + 2 \sqrt{2} }$