How do you factor $a^9 + b^12 c^15$ ?

Question

1874 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-05T15:27:25

$a^9+b^12c^15 = (a^3+b^4c^5)(a^6-a^3b^4c^5+b^8c^10)$

$a^9+b^12c^15$ is of the form $A^3+B^3$ where $A=a^3$ and $B=b^4c^5$ .

The sum of cubes identity tells us that:

$A^3+B^3 = (A+B)(A^2-AB+B^2)$

Hence:

$a^9+b^12c^15=(a^3)^3+(b^4c^5)^3$

$= (a^3+b^4c^5)((a^3)^2-(a^3)(b^4c^5)+(b^4c^5)^2)$

$= (a^3+b^4c^5)(a^6-a^3b^4c^5+b^8c^10)$

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

$= (a^3+b^4c^5)(a^3+omega b^4c^5)(a^3+omega^2 b^4c^5)$

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

How do you factor a^9 + b^12 c^15a^9 + b^12 c^15?