Question

If x-1/x=3; find the value of xcube+1/xcube
Plxx answer it!!!

Answer 1

$\underline{\underline{\Large{\mathfrak{Solution : }}}}$




$\underline{\textsf{We know that , }} \\ \\ \mathsf{\implies a^3 \: + \: b^3 \: = \: ( a \: + \: b )( a^2 \: + \: b^2 \: - \: ab ) }$




$\mathsf{Replace \: a \: by \: x \: and \: b \: by \: \dfrac{1}{x}.}$




$\mathsf{\implies x^3 \: + \: \left( \dfrac{1}{x} \right)^3 \: = \: \left( x \: + \: \dfrac{1}{x} \right) \left( x^2 \: + \: \dfrac{1}{x^2} \: - \: x \: \times \: \dfrac{1}{x} \right) } \\ \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \left(x \: + \: \dfrac{1}{x} \right) \left( {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 1 \right) \qquad...(1)}$





$\textsf{Now,} \\ \\ \mathsf{\implies x \: - \: \dfrac{1}{x} \: = \: 3 } \\ \\ \textsf{Squaring on both sides, } \\ \\ \mathsf{\implies \left( x \: - \: \dfrac{1}{x} \right)^2 \: = \: 3^2 } \\ \\ \mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \: = \: 9} \\ \\ \mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 2 \: = \: 9}$


$\mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 9 \: + \: 2 } \\ \\ \mathsf{ \implies { {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } } \: = \: 11 \: \qquad...(2)}$



$\underline{\mathsf{Now , add \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: to \: both \: sides , }}$




$\mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: + \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: = \: 11 \: + \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: }$

$\mathsf{ \implies \left(x \: + \: \dfrac{1}{ {x} } \right)^{2} \: = \: 11 \: + \: 2} \\ \\ \mathsf{ \implies \left(x \: + \: \dfrac{1}{ {x} } \right)^{2} \: = \: 13} \\ \\ \mathsf{ \implies x \: + \: \dfrac{1}{ {x} } \: = \pm\sqrt{13} \qquad...(3) }$



$\textsf{Plug the value of (2) and (3) in (1),} \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \left(x \: + \: \dfrac{1}{x} \right) \left( {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 1 \right) } \\ \\ \textsf{Plug the value of (2) and (3) in (1),} \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \pm \sqrt{13} \left( 11 \: - \: 1 \right) } \\ \\ \mathsf{ \: \: \therefore \: \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \pm10 \sqrt{13} }$

Answer 2

$\underline{\underline{\Large{\mathfrak{Solution : }}}} </p><p></p><p> </p><p> </p><p> </p><p>\\\\\\</p><p></p><p></p><p></p><p></p><p>\begin{lgathered}\underline{\textsf{We know that , }} \\ \\ \mathsf{\implies a^3 \: + \: b^3 \: = \: ( a \: + \: b )( a^2 \: + \: b^2 \: - \: ab ) }\end{lgathered} </p><p></p><p>\\\\\\</p><p></p><p></p><p></p><p>\mathsf{Replace \: a \: by \: x \: and \: b \: by \: \dfrac{1}{x}.}</p><p></p><p>\\\\\\</p><p></p><p></p><p>\begin{lgathered}\mathsf{\implies x^3 \: + \: \left( \dfrac{1}{x} \right)^3 \: = \: \left( x \: + \: \dfrac{1}{x} \right) \left( x^2 \: + \: \dfrac{1}{x^2} \: - \: x \: \times \: \dfrac{1}{x} \right) } \\ \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \left(x \: + \: \dfrac{1}{x} \right) \left( {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 1 \right) \qquad...(1)}\end{lgathered} </p><p></p><p> </p><p> </p><p>\\\\\\</p><p></p><p></p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{Now,} \\ \\ \mathsf{\implies x \: - \: \dfrac{1}{x} \: = \: 3 } \\ \\ \textsf{Squaring on both sides, } \\ \\ \mathsf{\implies \left( x \: - \: \dfrac{1}{x} \right)^2 \: = \: 3^2 } \\ \\ \mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \: = \: 9} \\ \\ \mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 2 \: = \: 9}\end{lgathered} </p><p>\\\\\\</p><p></p><p> </p><p></p><p></p><p>\begin{lgathered}\mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 9 \: + \: 2 } \\ \\ \mathsf{ \implies { {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } } \: = \: 11 \: \qquad...(2)}\end{lgathered} \\\\\\</p><p></p><p></p><p> </p><p> </p><p></p><p></p><p></p><p>\underline{\mathsf{Now , add \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: to \: both \: sides , }} \\\\\\</p><p></p><p></p><p> </p><p> </p><p></p><p></p><p></p><p></p><p>\mathsf{ \implies {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: + \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: = \: 11 \: + \: \left(2 \: \cdot \: x \: \cdot \: \dfrac{1}{x} \right) \: }\\\\\\</p><p></p><p></p><p>\begin{lgathered}\mathsf{ \implies \left(x \: + \: \dfrac{1}{ {x} } \right)^{2} \: = \: 11 \: + \: 2} \\ \\ \mathsf{ \implies \left(x \: + \: \dfrac{1}{ {x} } \right)^{2} \: = \: 13} \\ \\ \mathsf{ \implies x \: + \: \dfrac{1}{ {x} } \: = \pm\sqrt{13} \qquad...(3) }\end{lgathered} \\\\\\</p><p></p><p></p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{Plug the value of (2) and (3) in (1),} \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \left(x \: + \: \dfrac{1}{x} \right) \left( {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: - \: 1 \right) } \\ \\ \textsf{Plug the value of (2) and (3) in (1),} \\ \\ \mathsf{ \implies \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \pm \sqrt{13} \left( 11 \: - \: 1 \right) } \\ \\ \mathsf{ \: \: \therefore \: \: {x}^{3} \: + \: \dfrac{1}{ {x}^{3} } \: = \: \pm10 \sqrt{13} }\end{lgathered}$