Question

Factorise the following
$\begin{gathered} \sf{ {x}^{16} - {y}^{16} + {x}^{8} + {y}^{8} }\end{gathered}$
​

Answer 1

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

Given expression is

$\rm \: {x}^{16} - {y}^{16} + {x}^{8} + {y}^{8}$

can be rewritten as

$\rm \: = \: ( {x}^{16} - {y}^{16} )+({x}^{8} + {y}^{8})$

$\rm \: = \: \bigg[{( {x}^{8} )}^{2} - {( {y}^{8} )}^{2}\bigg] + ( {x}^{8} + {y}^{8})$

We know,

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

So, using this identity, we get

$\rm \: = \: ( {x}^{8} + {y}^{8})( {x}^{8} - {y}^{8}) + ( {x}^{8} + {y}^{8})$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[ {x}^{8} - {y}^{8} + 1\bigg]$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[ {( {x}^{4} )}^{2} - {( {y}^{4} )}^{2} + 1\bigg]$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[( {x}^{4} + {y}^{4})( {x}^{4} - {y}^{4}) + 1\bigg]$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[( {x}^{4} + {y}^{4})( {( {x}^{2})}^{2} - {( {y}^{2})}^{2}) + 1\bigg]$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[( {x}^{4} + {y}^{4})( {x}^{2} + {y}^{2})( {x}^{2} - {y}^{2})+ 1\bigg]$

$\rm \: = \: ( {x}^{8} + {y}^{8})\bigg[( {x}^{4} + {y}^{4})( {x}^{2} + {y}^{2})(x + y)(x - y)+ 1\bigg]$

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

Additional Information :-

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