Question

Please help in it is simple question5E%7B3%7D%20%20-%2027" id="TexFormula1" title=" {x}^{2} - 4 \\ {x}^{3} - 27" alt=" {x}^{2} - 4 \\ {x}^{3} - 27" align="absmiddle" class="latex-formula">
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Answer 1

x^2 - 4 = (x-2) x (x + 2)
x^3 - 27 = (x-3) x (x^2+ 3x+9)

Answer 2

1)

$\sf {x}^{2} - 4 \\ \sf \: Use \: the \: sum \: product \: method$

$\sf {x}^{2} - 4 \\ = \sf{x}^{2} + 2x - 2x - 4$

$\sf \: Now \: take \: the \: common \: factor \: out \\ \sf{x}^{2} + 2x - 2x - 4 \\\sf = x(x + 2) - 2(x + 2)$

$\sf \: Factorize \: it \\ \sf \: x(x + 2) - 2(x + 2) \\ = \sf(x - 2)(x + 2)$

Answer ⟶ $\boxed{\bf{(x-2)(x+2)}}$

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2)

$\sf {x}^{3} - 27$

$\sf {x}^{3} \: and \: 27 \: ( {3}^{3} ) \: are \: perfect \: real \: cubes.$

$\sf \: So \: use \: the \: algebraic \: identity \: \\ \sf {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

$\sf \: a = \sqrt[3]{x^{3}} = x \\ \sf \: b = \sqrt[3]{27} = 3$

$\sf \: {x}^{3} - {3}^{3} \\ \sf= (x - 3)( {x}^{2} + 3x + {3}^{2} ) \\ = \sf \: (x - 3)( {x}^{2} + 3x + 9)$

Answer ⟶ $\boxed{\bf{(x-3)(x^{2}+3x+9)}}$