Question

$\large\red{\sf{Question}}$

$\sf{One \: of \: the \: factors \: of \: {x}^{2} + \frac{1}{ {x}^{2} } + 2 - 2x - \frac{2}{x} }$
$\large\underline{\sf{A \rightarrow x - \frac{1}{x} }}$
$\large\underline{\sf{B \rightarrow x + \frac{1}{x} - 1}}$
$\large\underline{\sf{C \rightarrow x + \frac{1}{x} }}$
$\large\underline{\sf{D \rightarrow {x}^{2} + \frac{1}{ {x}^{2} } }}$
​

Answer 1

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

Given expression is

$\rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 - 2x - \dfrac{2}{x}$

can be re-arranged as

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

can be further rewritten as

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

We know,

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

So, using this, above expression can be rewritten as

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

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

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

So,

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

Hence, option C is correct.

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

Additional Information :-

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