# How do you factor completely #64 - 27y^3 #?

##### 1 Answer

Aug 5, 2016

#### Explanation:

The difference of cubes identity can be written:

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

Note that

#64-27y^3#

#=4^3-(3y)^3#

#= (4-y)(4^2+4y+y^2)#

#= (4-y)(16+4y+y^2)#

The remaining quadratic has negative discriminant. It cannot be factored further with Real coefficients. If we allow Complex coefficients then we can factor it completely:

#= (4-y)(4-omega y)(4-omega^2 y)#

where