Find the value ofif when

1.	Find the value ofif when
Answer» $\large\underline{\bold{Given \:Question - }}$ $\sf \: Find \: the \: value \: of \: (<klux>X</klux>^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - 1) \: when \: x = <klux>8</klux>.$ $\large\underline{\bold{Solution-}}$ We know that $\bf \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$ $\tt \: Put \: x \: = \: {x}^{\frac{1}{3} } \: \: and \: \: y \: = \: {x}^{ - \frac{1}{3} } , \: we \: get$ $\bf \: {\bigg( {x}^{\frac{1}{3}}\bigg) }^{3} \:+\:{\bigg( {x}^{ \frac{ - 1}{3} } \bigg) }^{3} = (x^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - x^ \frac{1}{3} x^{ - \frac{1 }{3}} )$ $\bf \: x + {x}^{ - 1} = \sf \: (x^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - 1) \:$ $\bf \: <klux>HENCE</klux>, \: (x^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - 1) \: = x + \dfrac{1}{x}$ Put x = 8, we get $\therefore \: \sf \: x + \dfrac{1}{x} = 8 + \dfrac{1}{8} = \dfrac{64 + 1}{8} = \dfrac{65}{8}$ Hence, The value of $\sf \:\: (x^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - 1) \:, \: when \: x = 8 \: is \: \dfrac{65}{8}$ ADDITIONAL IMPORTANT Identities :- $\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$ $\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$ $\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}$ $\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$ $\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} }}$ $\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + {3y}^{2} x - {y}^{3} }}$ $\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy}}$ $\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}$

Answer»

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Find \: the \: value \: of \: (<klux>X</klux>^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - 1) \: when \: x = <klux>8</klux>.$

$\large\underline{\bold{Solution-}}$

We know that

$\bf \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

$\tt \: Put \: x \: = \: {x}^{\frac{1}{3} } \: \: and \: \: y \: = \: {x}^{ - \frac{1}{3} } , \: we \: get$

$\bf \: {\bigg( {x}^{\frac{1}{3}}\bigg) }^{3} \:+\:{\bigg( {x}^{ \frac{ - 1}{3} } \bigg) }^{3} = (x^ \frac{1}{3} + x ^\frac{ - 1}{3} )(x ^\frac{2}{3} + x ^\frac{ - 2}{3} - x^ \frac{1}{3} x^{ - \frac{1 }{3}} )$