How do you factor $m^3 - n^3$ ?

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Answer 1 · 2015-12-30T18:44:13

$m^3-n^3 = (m-n)(m^2+mn+n^2)$

This is a standard identity known as the "difference of cubes" identity.

The remaining quadratic factor can only be factored further using Complex coefficients:

$(m^2+mn+n^2) = (m- omega n)(m - omega^2 n)$

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

How do you factor m^3 - n^3m^3 - n^3?