Question

${a}^{2} - (b - c) {}^{2}$
​

Answer 1

Answer:

Answer:

a²-(b-c)² = (a+b-c)(a-b+c)

Explanation:

Factorising a²-(b-c)²

we know the algebraic identity:

x² - y² = (x + y) (x − y)

=[a+(b-c)][a-(b-c)]

= (a+b-c)(a-b+c)

Therefore,

a²-(b-c)2(a+b-c)(a-b+c)

Answer 2

Answer:

$\tt\large{\textbf{{{\color{navy}{ \: \: \: \: \: \:Ꭺŋ}}{\red{ѕw}}{\pink{єr♤࿐}}{\color{pink}{:}}}}}$

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$\tt\large{ \pink}{1. \: \: expand \: the \: square : - }$

${ \tt{ \pink}{a}^{2} } - 1(b - c) {}^{2}$

$\tt{a}^{2} - 1(b - c)(b - c)$

$\tt\large{2. \: \: distribute : - }$

$\tt {a}^{2} - 1(b - c )( b - c )$

$\tt{a}^{2} - 1(b(b -c ) - c(b - c ))$

$\tt{3. \: distribute : - \: }$

$\tt{a}^{2} - 1(b(b - c) - c(b - c))$

$\tt {a}^{2} - 1( {b}^{2} - bc - c(b - c))$

$\tt\large{4. \: distribute : - \: }$

$\tt{a}^{2} - 1( {b}^{2} - bc - c(b - c))$

$\tt {a}^{2} - 1( {b}^{2} - bc - bc + {c}^{2} )$

$\tt\large{5. \: combine \: like \: terms : - \: }$

$\tt {a}^{2} - 1( {b}^{2} - bc - bc + {c}^{2} )$

$\tt {a}^{2} - 1( {b}^{2} - 2bc + {c}^{2} )$

$\tt\large{6. \: distribute : - }$

$\tt {a}^{2} - 1( {b}^{2} - 2bc + {c}^{2} )$

$\tt {a}^{2} - 1 {b}^{2} + 2bc - 1 {c}^{2}$

$\tt\large{ \: solution = }$

$\tt {a}^{2} - 1 {b}^{2} + 2 - 1 {c}^{2}$

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