Question

Factorize a²+2ab+b²-x²+2xy-y²​

Answer 1

using identity (a+b) =a sq. + 2ab + b sq.

= (a+b)sq. - (x+y)sq.

now, using identity a sq. - b sq. = (a+b)(a- b)

(a+b+x+y)(a+b-x+y) = ANSWER

Answer 2

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

Given expression is

$\rm \: {a}^{2} + 2ab + {b}^{2} - {x}^{2} + 2xy - {y}^{2}$

can be regrouped as

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

We know,

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

and

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

So, using these results, we get

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

We know,

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

So, using this identity, we get

$\rm \: = \: [(a + b) + (x - y)] \: [(a + b) - (x - y)]$

$\rm \: = \: (a + b + x - y) \: (a + b - x + y)$

Hence,

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

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

Additional Information :-

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