How do you factor completely $mn^4 + m^4n$ ?

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Answer 1 · 2015-12-04T15:19:55

Separate out the common factor $mn$ then use the sum of cubes identity to find:

$mn^4+m^4n = mn(m+n)(m^2-mn+n^2)$

The sum of cubes identity can be written:

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

We can separate out the common factor $mn$ then use the sum of cubes identity with $a=m$ and $b=n$ to find:

$mn^4+m^4n = mn(n^3+m^3)$

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

If we allow Complex coefficients then this can be factored further as:

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

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

How do you factor completely mn^4 + m^4nmn^4 + m^4n?