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

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Answer 1 · 2016-03-15T14:45:31

Use the difference of cubes identity to find:

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

Both #x^3# and #27=3^3# are perfect cubes. So we can use the difference of cubes identity:

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

with #a=x# and #b=3# as follows:

#x^3-27#

#=x^3-3^3#

#=(x-3)(x^2+x(3) + 3^2)#

#=(x-3)(x^2+3x+9)#

This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:

#=(x-3)(x-3omega)(x-3omega^2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.