Question

Find the value of
$(x \frac{1}{3} + x \frac{ - 1}{3} )(x \frac{2}{3} + x \frac{ - 2}{3} - 1)$
if when
$x = 8$
​

Answer 1

Answer:

0

Explanation:

Solution is in picture

Answer 2

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

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

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

We know that

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

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

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

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

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

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 :-

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

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

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

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

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

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

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

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}$