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#).