Question

Please help in it is simple question

`{x}^(2) - 4 \\ {x}^(3) - 27`
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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)}}`