Answer:

use identity

(a + b) (a - b) = a² - b²

Verified answer

Answer:

[tex]\qquad\boxed{ \sf{ \: ( \sqrt{a + b} + \sqrt{c}) \: ( \sqrt{a + b} - \sqrt{c}) = a + b - c \: }} \\ \\ [/tex]

Step-by-step explanation:

Given algebraic expression is

[tex]\sf \: ( \sqrt{a + b} + \sqrt{c}) \: ( \sqrt{a + b} - \sqrt{c}) \\ \\ [/tex]

We know,

[tex]\qquad\boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:x= \sqrt{a + b} \qquad \: \\ \\& \qquad \:\sf \: y= \sqrt{c}\end{aligned}} \qquad \: \\ \\ [/tex]

So, on substituting the values in above formula, we get

[tex]\sf \: = \: {( \sqrt{a + b} )}^{2} - {( \sqrt{c})}^{2} \\ \\ [/tex]

[tex]\sf \: = \: a + b - c \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: ( \sqrt{a + b} + \sqrt{c}) \: ( \sqrt{a + b} - \sqrt{c}) = a + b - c \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\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}[/tex]