How do you factor $x^3-27$ ?

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Answer 1 · 2015-06-03T10:09:47

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

Therefore :

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

To convince ourself, we can use Horner's scheme :

Firstly, we need to find a $x$ -value for which $x^3-27 = 0$ .
Here, it's pretty obvious : $x = 3$ .

Then, we can use the scheme :

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Therefore, $x^3-27 = (x-3)(x^2+3x+9)$ .

(We can't factorise $x^2 + 3x + 9$ , because $Delta = b^2-4ac = 9 - 4*1*9 = -27 < 0$ ).

How do you factor x^3-27x^3-27?