How do you factor $e^6 + f^3$ ?

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Answer 1 · 2015-12-09T19:03:21

Use the sum of cubes identity to find:

$e^6+f^3 = (e^2+f)(e^4-e^2f+f^2)$

The sum of cubes identity may be written:

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

If we let $a=e^2$ and $b=f$ then we find:

$e^6+f^3 = (e^2)^3+f^3$

$=(e^2+f)((e^2)^2-(e^2)f+f^2)$

$=(e^2+f)(e^4-e^2f+f^2)$

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

$=(e^2+f)(e^2+omega f)(e^2+omega^2 f)$

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

How do you factor e^6 + f^3e^6 + f^3?