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

Question

22105 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#.