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#.