Question

a² + 2ab + b² – 16
Factorise these values

Answer 1

$\large\star{\boxed{\rm{\purple{ (a+b)² \ = \ a² \ + \ b² \ + \ 2ab}}}}$

So, a² + 2ab + b² – 16 can be written as (a + b)² - 16.

$\tt{↬\ (a + b)² \ - \ 16}$

Answer 2

$\large\underline\blue{\bold{Given \:  Question }}$

$\rm : \implies \:Factorise \: {a}^{2} + 2ab + {b}^{2} - 16$

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$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed {\pink{\rm : \implies \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }}$

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

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$\large\underline\purple{\bold{Solution :- }}$

$\rm : \implies \:{a}^{2} + 2ab + {b}^{2} - 16$

$\rm : \implies \:({a}^{2} + 2ab + {b}^{2}) - 16$

$\rm : \implies \: {(a + b)}^{2} - {(4)}^{2}$

$\rm : \implies \:(a + b + 4)(a + b - 4)$

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Explore more :-

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

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

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

$\boxed {\pink{\rm \: (a + b + c)^{2} = {a}^{2} + {b}^{2} + {c}^{2} +2ab + 2bc + 2ca }}$

$\boxed {\pink{\rm \: (a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) }}$

$\boxed {\pink{\rm \: (a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) }}$

$\boxed {\pink{\rm \: a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) }}$

$\boxed {\pink{\rm \: a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab) }}$