Question

Simplify
[x^(1/3)-x^(-1/3)]*[x^(2/3)+1+x^(-2/3)]

Answer 1

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

Given expression is

$\bigg[{\bigg(x\bigg) }^{\dfrac{1}{3} } - {\bigg(x\bigg) }^{ - \dfrac{1}{3} }\bigg]\bigg[{\bigg(x\bigg) }^{\dfrac{2}{3} } + 1 + {\bigg(x\bigg) }^{ - \dfrac{2}{3} }\bigg]$

can be rewritten as

$\rm \: = \bigg[{\bigg(x\bigg) }^{\dfrac{1}{3} } - {\bigg(x\bigg) }^{ - \dfrac{1}{3} }\bigg]\bigg[{\bigg(x\bigg) }^{\dfrac{2}{3} } +{\bigg(x\bigg) }^{\dfrac{1}{3} }{\bigg(x\bigg) }^{ - \dfrac{1}{3} } + {\bigg(x\bigg) }^{ - \dfrac{2}{3} }\bigg]$

Let we assume that

$\rm \: {\bigg(x\bigg) }^{\dfrac{1}{3} } = a$

and

$\rm \: {\bigg(x\bigg) }^{ - \dfrac{1}{3} } = b$

So, above expression can be rewritten as

$\rm \: = \: (a - b)( {a}^{2} + ab + {b}^{2})$

$\rm \: = \: {a}^{3} - {b}^{3}$

$\rm \: = \: {\bigg(x\bigg) }^{\dfrac{1}{3} \times 3} - {\bigg(x\bigg) }^{ - \dfrac{1}{3} \times 3}$

$\rm \: = \: x - {x}^{ - 1}$

$\rm \: = \: x - \dfrac{1}{x}$

Hence,

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

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$