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emmas19 · Jan 21, 2011, 11:59 AM

how do you factor m^6-27n^3?
or m^6+27n^3?

ebaines · Jan 21, 2011, 12:25 PM

First notice that 27 = 3^3. So these equations can be rewritten as the difference of cubes for the first one, and the sum of cubes for the second:

$ m^6 -27n^3 = (m^2)^3-(3n)^3 $

and

$ m^6 +27n^3 = (m^2)^3+(3n)^3 $

The difference of cubes can be factored like this:

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

and the sum of cubes can be factored like this:

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

Hope this helps.