How do you factor completely $27x^3-125y^3$ ?

Question

22504 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-26T21:30:50

$27x^3-125y^3=(3x-5y)(9x^2+15xy+25y^2)$

Notice that both of the terms are perfect cubes:

$27x^3 = (3x)^3$

$125y^3 = (5y)^3$

So we can conveniently use the difference of cubes identity:

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

with $A=3x$ and $B=5y$ as follows:

$27x^3-125y^3$

$=(3x)^3-(5y)^3$

$=(3x-5y)((3x)^2+(3x)(5y)+(5y)^2)$

$=(3x-5y)(9x^2+15xy+25y^2)$

The remaining quadratic factor has no linear factors with Real coefficients. If we allow Complex coefficients then we can factor a little further:

$=(3x-5y)(3x-5omega y)(3x-5 omega^2 y)$

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

How do you factor completely 27x^3-125y^327x^3-125y^3?