How do you factor #27x^3-512#?

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Answer 1 · 2015-12-17T20:13:32

Use the difference of cubes identity to find:

#27x^3-512 = (3x-8)(9x^2+24x+64)#

The difference of cubes identity may be written:

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

Notice that #27x^3 = (3x)^3# and #512 = 8^3# are both perfect cubes. So let #a=3x# and #b=8# to find:

#27x^3-512#

#=(3x)^3-8^3#

#=(3x-8)((3x)^2+(3x)(8)+8^2)#

#=(3x-8)(9x^2+24x+64)#

If you allow Complex coefficients then this factors a little further:

#=(3x-8)(3x-8omega)(3x-8omega^2)#

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