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

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Answer 1 · 2015-12-30T18:44:13

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

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

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

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

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

How do you factor m^3 - n^3m3−n3m^3 - n^3?