How do you factor #1000x^3+27#?

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Answer 1 · 2016-08-28T20:18:39

#1000x^3+27=(10x+3)(100x^2-30x+9)#

The sum of cubes identity can be written:

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

Use this with #a=10x# and #b=3# as follows:

#1000x^3+27#

#=(10x)^3+3^3#

#=(10x+3)((10x)^2-(10x)(3)+3^2)#

#=(10x+3)(100x^2-30x+9)#

This is as far as we can go with Real coefficients. If we allow Complex coefficients then it can be factored further as:

#=(10x+3)(10x+3omega)(10x+3omega^2)#

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